case study questions for class 12 physics chapter 1

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CBSE Class 12 Physics Case Study Questions PDF Download

Case Study questions for the Class 12 Physics board exams are available here. You can read the Class 12 Physics Case Study Questions broken down by chapter. Subject matter specialists and seasoned teachers created these quizzes. You can verify the right response to each question by referring to the answer key, which is also provided. To achieve success on your final exams, practice the following questions.

CBSE (Central Board of Secondary Education) is a renowned educational board in India that designs the curriculum for Class 12 Physics with the goal of promoting a scientific temperament and nurturing critical thinking among students. As part of their Physics examination, CBSE includes case study questions to assess students’ ability to apply theoretical knowledge to real-world scenarios effectively.

Chapter-wise Solved Case Study Questions for Class 12 Physics

Before the exams, students in class 12 should review crucial Physics Case Study issues. They will gain a better understanding of the kinds of Case Study questions that may be asked in Physics exams for Grade 12. These questions were created by our highly qualified standard 12 Physics staff based on the questions that appeared most frequently in last year’s exams. The solutions have been written in a way that will make them simple to grasp and will aid students in grade 12 in understanding the topics.

Best Books for Class 12 Physics Exams

Strictly in accordance with the new term-by-term curriculum for the class 12 board exams to be held in the academic year 2023–2024, which will include multiple choice questions based on new board typologies including stand-alone MCQs and case-based MCQs with an assertion–reason. Included are inquiries from the official CBSE Question Bank that was released in April 2024. What changes have been made to the book: strictly in accordance with the term-by-term syllabus for the board exams that will be held during the 2024 academic year? Chapter- and topic-based Questions with multiple choices that are based on the unique evaluation method used for the Class 12th Physics Final Board Exams.

Key Benefits of Solving CBSE Class 12 Physics Case Study Questions

• Application of Concepts: Case study questions demand the application of theoretical knowledge in practical scenarios, preparing students for real-world challenges and professional pursuits.
• Critical Thinking: By evaluating and analyzing case studies, students develop critical thinking abilities, enabling them to approach complex problems with a logical mindset.
• In-Depth Understanding: Addressing case study questions necessitates a thorough understanding of physics concepts, leading to a more profound comprehension of the subject matter.
• Holistic Evaluation: CBSE adopts case study questions as they provide a holistic evaluation of a student’s aptitude and proficiency in physics, moving beyond rote memorization.
• Preparation for Competitive Exams: Since competitive exams often include similar application-based questions, practicing case study questions equips students for various entrance tests.

How to Approach CBSE Class 12 Physics Case Study Questions

• Read and Analyze Thoroughly: Carefully read the case study to grasp its context and identify the underlying physics principles involved.
• Identify Relevant Concepts: Highlight the physics theories and concepts applicable to the given scenario.
• Create a Systematic Solution: Formulate a step-by-step solution using the identified concepts, explaining each step with clarity.
• Include Diagrams and Charts: If relevant, incorporate diagrams, charts, or graphs to visually represent the situation, aiding better comprehension.
• Double-Check Answers: Always review your answers for accuracy, ensuring that they align with the principles of physics.

Tips for Excelling in CBSE Class 12 Physics

• Conceptual Clarity: Focus on building a strong foundation of physics concepts, as this will enable you to apply them effectively to case study questions.
• Practice Regularly: Dedicate time to solving case study questions regularly, enhancing your proficiency in handling real-world scenarios.
• Seek Guidance: Don’t hesitate to seek guidance from teachers, peers, or online resources to gain additional insights into challenging concepts.
• Time Management: During exams, practice efficient time management to ensure you allocate enough time to each case study question without rushing.
• Stay Positive: Approach case study questions with a positive mindset, embracing them as opportunities to showcase your skills and knowledge.

CBSE Class 12 Physics case study questions play a pivotal role in promoting practical understanding and critical thinking among students. By embracing these questions as opportunities for growth, students can excel in their physics examinations and beyond.

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1569+ Class 12 Physics Case Study Questions PDF Download

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So, you’re in Class 12 and the concept of Physics case study questions is beginning to loom large. Are you feeling a little lost? Fear not! This article will guide you through everything you need to know to conquer these challenging yet rewarding Class 12 Physics Case Study Questions.

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

We have provided here Case Study questions for the Class 12 Physics for board exams. You can read these chapter-wise Case Study questions. These questions are prepared by subject experts and experienced teachers. The answer and solutions are also provided so that you can check the correct answer for each question. Practice these questions to score excellent marks in your final exams.

Physics is more than just formulas and equations, right? It’s a way to interpret the natural world, and case studies provide a perfect opportunity for students to apply theoretical knowledge to real-world situations. So, what’s the importance of case studies in class 12 Physics? Well, they help you to solidify your understanding and enhance your analytical skills, which are invaluable for exams and beyond.

Table of Contents

Case Study-Based Questions for Class 12 Physics

There are total of 14 chapters Electric Charges And Fields, Electrostatic Potential And Capacitance, Current Electricity, Moving Charges And Magnetism, Magnetism And Matter, Electromagnetic Induction, Alternating Current, Electromagnetic Waves, Ray Optics and Optical Instruments, Wave Optics, Dual Nature Of Radiation And Matter, Atoms, Nuclei, Semiconductor Electronics Materials Devices, And Simple Circuits.

• Case Study Based Questions on class 12 Physics Chapter 1 Electric Charges And Fields
• Case Study Based Questions on Class 12 Physics Chapter 2 Electrostatic Potential And Capacitance
• Case Study Based Questions on Class 12 Physics Chapter 3 Current Electricity
• Case Study Based Questions on Class 12 Physics Chapter 4 Moving Charges And Magnetism
• Case Study Based Questions on Class 12 Physics Chapter 5 Magnetism And Matter
• Case Study Based Questions on Class 12 Physics Chapter 6 Electromagnetic Induction
• Case Study Based Questions on Class 12 Physics Chapter 7 Alternating Current
• Case Study Based Questions on Class 12 Physics Chapter 8 Electromagnetic waves
• Case Study Based Questions on Class 12 Physics Chapter 9 Ray Optics and Optical Instruments
• Case Study Based Questions on class 12 Physics Chapter 10 Wave Optics
• Case Study Based Questions on class 12 Physics Chapter 11 Dual Nature of Matter and Radiation
• Case Study Based Questions on class 12 Physics Chapter 12 Atoms
• Case Study Based Questions on class 12 Physics Chapter 13 Nuclei
• Case Study Based Questions on class 12 Physics Chapter 14 Semiconductor Electronics
• Class 12 Chemistry Case Study Questions
• Class 12 Biology Case Study Questions
• Class 12 Maths Case Study Questions

Class 12 students should go through important Case Study problems for Physics before the exams. This will help them to understand the type of Case Study questions that can be asked in Grade 12 Physics examinations. Our expert faculty for standard 12 Physics have designed these questions based on the trend of questions that have been asked in last year’s exams. The solutions have been designed in a manner to help the grade 12 students understand the concepts and also easy-to-learn solutions.

How to Approach Physics Case Study Questions

Ah, the million-dollar question! How do you tackle these daunting case study questions? Let’s dive in.

Effective Reading Techniques

A vital part of cracking case study questions is understanding the problem correctly. Don’t rush through the question, take your time to fully grasp the scenario, and don’t skip over the details – they can be critical!

Conceptual Analysis

Once you’ve understood the problem, it’s time to figure out which physics principles apply. Remember, your foundational knowledge is your greatest weapon here.

Solving Strategies

Next, put pen to paper and start solving! Follow a systematic approach, check your calculations, and make sure your answer makes sense in the context of the problem.

Best Books for Class 12 Physics Exams

Class 12 Physics Syllabus 2024

Chapter–1: Electric Charges and Fields

Electric charges, Conservation of charge, Coulomb’s law-force between two point charges, forces between multiple charges; superposition principle, and continuous charge distribution.

Electric field, electric field due to a point charge, electric field lines, electric dipole, electric field due to a dipole, torque on a dipole in a uniform electric field.

Electric flux, statement of Gauss’s theorem and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet, and uniformly charged thin spherical shell (field inside and outside).

Chapter–2: Electrostatic Potential and Capacitance

Electric potential, potential difference, electric potential due to a point charge, a dipole and system of charges; equipotential surfaces, electrical potential energy of a system of two-point charges and of electric dipole in an electrostatic field.

Conductors and insulators, free charges and bound charges inside a conductor. Dielectrics and electric polarization, capacitors and capacitance, combination of capacitors in series and in parallel, capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor (no derivation, formulae only).

Chapter–3: Current Electricity

Electric current, flow of electric charges in a metallic conductor, drift velocity, mobility and their relation with electric current; Ohm’s law, V-I characteristics (linear and non-linear), electrical energy and power, electrical resistivity and conductivity, temperature dependence of resistance, Internal resistance of a cell, potential difference and emf of a cell, combination of cells in series and in parallel, Kirchhoff’s rules, Wheatstone bridge.

Chapter–4: Moving Charges and Magnetism

Concept of magnetic field, Oersted’s experiment.

Biot – Savart law and its application to current carrying circular loop.

Ampere’s law and its applications to infinitely long straight wire. Straight solenoid (only qualitative treatment), force on a moving charge in uniform magnetic and electric fields.

Force on a current-carrying conductor in a uniform magnetic field, force between two parallel current-carrying conductors-definition of ampere, torque experienced by a current loop in uniform magnetic field; Current loop as a magnetic dipole and its magnetic dipole moment, moving coil galvanometerits current sensitivity and conversion to ammeter and voltmeter.

Chapter–5: Magnetism and Matter

Bar magnet, bar magnet as an equivalent solenoid (qualitative treatment only), magnetic field intensity due to a magnetic dipole (bar magnet) along its axis and perpendicular to its axis (qualitative treatment only), torque on a magnetic dipole (bar magnet) in a uniform magnetic field (qualitative treatment only), magnetic field lines.

Magnetic properties of materials- Para-, dia- and ferro – magnetic substances with examples, Magnetization of materials, effect of temperature on magnetic properties.

Chapter–6: Electromagnetic Induction

Electromagnetic induction; Faraday’s laws, induced EMF and current; Lenz’s Law, Self and mutual induction.

Chapter–7: Alternating Current

Alternating currents, peak and RMS value of alternating current/voltage; reactance and impedance; LCR series circuit (phasors only), resonance, power in AC circuits, power factor, wattless current. AC generator, Transformer.

Chapter–8: Electromagnetic Waves

Basic idea of displacement current, Electromagnetic waves, their characteristics, their transverse nature (qualitative idea only). Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays) including elementary facts about their uses.

Chapter–9: Ray Optics and Optical Instruments

Ray Optics:  Reflection of light, spherical mirrors, mirror formula, refraction of light, total internal reflection and optical fibers, refraction at spherical surfaces, lenses, thin lens formula, lens maker’s formula, magnification, power of a lens, combination of thin lenses in contact, refraction of light through a prism.

Optical instruments:  Microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers.

Chapter–10: Wave Optics

Wave optics: Wave front and Huygen’s principle, reflection and refraction of plane wave at a plane surface using wave fronts. Proof of laws of reflection and refraction using Huygen’s principle. Interference, Young’s double slit experiment and expression for fringe width (No derivation final expression only), coherent sources and sustained interference of light, diffraction due to a single slit, width of central maxima (qualitative treatment only).

Chapter–11: Dual Nature of Radiation and Matter

Dual nature of radiation, Photoelectric effect, Hertz and Lenard’s observations; Einstein’s photoelectric equation-particle nature of light.

Experimental study of photoelectric effect

Matter waves-wave nature of particles, de-Broglie relation.

Chapter–12: Atoms

Alpha-particle scattering experiment; Rutherford’s model of atom; Bohr model of hydrogen atom, Expression for radius of nth possible orbit, velocity and energy of electron in his orbit, of hydrogen line spectra (qualitative treatment only).

Chapter–13: Nuclei

Composition and size of nucleus, nuclear force

Mass-energy relation, mass defect; binding energy per nucleon and its variation with mass number; nuclear fission, nuclear fusion.

Chapter–14: Semiconductor Electronics: Materials, Devices and Simple Circuits

Energy bands in conductors, semiconductors and insulators (qualitative ideas only) Intrinsic and extrinsic semiconductors- p and n type, p-n junction

Semiconductor diode – I-V characteristics in forward and reverse bias, application of junction diode -diode as a rectifier.

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You can click on the chapter-wise Case Study links given above and read the Class 12th Physics Case Study questions and answers provided by us. If you face any issues or have any questions please put your questions in the comments section below. Our teachers will be happy to provide you with answers.

FAQs Class 12 Physics Case Studies Questions

What is the best website for a  case   study  of physics  class   12 .

studyrate.in is the best website for Class 12 Physics Case Study Questions for Board Exams. Here you can find various types of Study Materials, Ebooks, Notes, and much more free of cost.

What are Physics case study questions?

Physics case study questions involve applying physics principles to solve real-world scenarios or problems.

How can I prepare for Physics case study questions?

Strengthen your basic concepts, practice solving a variety of problems, and make use of resources like the Class 12 Physics Case Study Questions PDF.

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myCBSEguide

• Case Study Questions Class...

Case Study Questions Class 12 Physics

Table of Contents

myCBSEguide App

Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

CBSE will ask two Case Study Questions in the CBSE class 12 Physics questions paper. Question numbers 15 and 16 are case-based questions where 5 MCQs will be asked based on a paragraph. Each theme will have five questions and students will have a choice to attempt any four of them. You can download CBSE Class 12 Physics case study questions from the myCBSEguide App and our free student dashboard .

You can Score High

CBSE class 12 Physics question paper will carry questions for 70 marks. Certainly, the question paper is a bit easier this year. It is because the syllabus is already reduced and there are more internal choices. Besides this, the case study questions are a plus to winning the game with good marks.

In simple words, all circumstances are in favour of the sincere students who are working hard to score high this year. Although it has been a difficult time for students as they were not getting the personal attention of the teachers. We know that online classes have their own limits, but we still expect better scores, especially from students who are putting extra effort into their studies.

Class 12 Physics Case Study Questions

CBSE class 12 Physics question paper will have case study questions too. These case-based questions will be objective type in nature. So, class 12 Physics students must prepare themselves for such questions. First of all, you should study NCERT Textbooks line by line and then you should practice as many questions as possible.

Case Study Syllabus

We know that CBSE has reduced the syllabus. Hence, practice only relevant questions. don’t waste time on case study questions from deleted portion. It is of no use. You can download the latest class 12 Physics case study questions from the myCBSEguide App.

Physics Case Studies

Class 12 Physics has many chapters but all chapters are not important for case studies. As we know case studies are not exactly given from NCERT books but these may be extracted from some newspaper articles, magazines, journals or other books. So, it is very much important that you are studying only the most relevant case studies. Here, the myCBSEguide app helps you a lot. We have case study questions that are prepared by a team of expert teachers. These experts exactly know what types of questions can come in exams.

Case Study Questions

There are a number of study apps available over the internet. But if you are a CBSE student and willing to get an app for the CBSE curriculum, you have very limited options. And if you want an app that is specifically designed for CBSE students, your search will definitely end on finding myCBSEguide. Case study questions are the latest updates in CBSE syllabus. It is exclusively available in the myCBSEguide app.

Here are some example questions. For more questions, you can download the myCBSEguide App.

Physics Case Study -1

Read  the following source and answer any four out of the following questions: Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of charges positive and negative charges. Also, like charges repel each other whereas unlike charges attract each other.

• -3.2  × ×  10 -18  C
• 3.2  × ×  10  18  C
• -3.2  × ×  10 -17  C
• 3.2  × ×  10  -17   C
• -1.6  × ×  10 -18  C
• 1.6  × ×  10  -18  C
• 2.6  × ×  10 -18  C
• 1.6  × ×  10 -21  C
• 9.1  × ×  10 -31  kg
• 9.1  × ×  10 -31  g
• 1.6  × ×  10 -19  kg
• 1.6  × ×  10 -19  g
• there is only a positive charge in the body
• there is positive as well as negative charge in the body but the positive charge is more than the negative charge
• there is equally positive and negative charge in the body but the positive charge lies in the outer regions
• the negative charge is displaced from its position
• valence electrons only
• electrons of inner shells
• both valence electrons and electrons of the inner shell.
• none of the above

Physics Case Study -2

Read the following source and answer any four out of the following questions: Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms. Also, Resistivity is the electrical resistance of a conductor of unit cross-sectional area, and unit length. … A characteristic property of each material, resistivity is useful in comparing various materials on the basis of their ability to conduct electric currents.

• nature of material
• temperature
• dimensions of material
• cross-sectional area
• length of wire
• wire’s nature
• all of the above
• more resistance
• less resistance
• same resistance

Physics Case Study -3

Read the source given below and answer any four out of the following questions: The Bohr model of the atom was proposed by Neil Bohr in 1915. It came into existence with the modification of Rutherford’s model of an atom. Rutherford’s model introduced the nuclear model of an atom, in which he explained that a nucleus (positively charged) is surrounded by negatively charged electrons.

• The energy of the electrons in the orbit is quantized
• The electron in the orbit nearest the nucleus has the lowest energy
• Electrons revolve in different orbits around the nucleus
• The position and velocity of the electrons in the orbit cannot be determined simultaneously
• Single proton
• Multiple electrons
• emitted only
• absorbed only
• both (a) and (b)
• none of these
•  It must emit a continuous spectrum
•  It loses its energy
• Gaining its energy
• A discrete spectrum
• dequantized

Physics Case Study & myCBSEguide App

We at myCBSEguide provide the best case study questions for CBSE class 12 Physics. We have Physics case study questions for every chapter in 12th class Physics. Students can access the Physics case study questions with answers on the myCBSEguide App or on the student dashboard . Here are some features that make myCBSEguide the best learning app for CBSE students:

• Updated syllabus
• Up to date question bank
• Model papers and 10-year questions
• NCERT and Exemplar sulutions
• Best quality learning videos
• Detailed revision notes

12 Physics Question Paper Design

Here is the question paper design for CBSE class 12 Physics. It shows the typology of the questions and their weightage in CBSE board exams.

QUESTION PAPER DESIGN Theory (Class: 12)

Maximum Marks: 70 Duration: 3 hrs

Note:  The above template is only a sample. Suitable internal variations may be made for generating similar templates keeping the overall weightage to different forms of questions and typology of questions the same.

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Electric Charges and Fields Class 12 Important Questions with Solutions

• Class 12 Important Question

Electric Charges and Fields Physics Class 12 Chapter 1 Important Questions PDF

Science is a complicated and challenging subject, as it involves a lot of theories and concepts that are hard to remember. Due to these complications, many students don't opt for science after finishing their 10th board exams. Those who opt for science face many challenges and have to work hard for scoring good marks in the exams. Class 12 is the most difficult.

Several of the chapters that students learn in class 12 are difficult to grasp. Similarly, class 12 science chapter 1 is regarded as one of the most challenging chapters. Students will study about electric charges and fields in this chapter. This chapter's pictures and ideas make it more convoluted and confusing. The greatest answer to this problem is to practise Electric Charges and Fields Important Questions with Answers PDF in order to clarify your concerns.

Download CBSE Class 12 Physics Important Questions 2024-25 PDF

Also, check CBSE Class 12 Physics Important Questions for other chapters along with Physics Class 12 Chapter 1 Important Questions PDF.

Related Chapters

Boost Your Performance in CBSE Exam Chapter 1 with Physics Class 12 Chapter 1 Important Questions PDF

Very Short Answer Questions – 1 Marks

1. Does the Force Between Two Point Charges Change if the Dielectric Constant of the Medium in Which they are Kept is Increased?

Ans:  Dielectric constant of a medium is given by 

$k=\frac{{{F}_{V}}}{{{F}_{M}}}=\frac{\text{force between the charges in vaccum}}{\text{force between two charges in medium}}$

$\Rightarrow {{F}_{M}}=\frac{{{F}_{V}}}{k}$

From the above expression, it is clear that as $k$ is increased, ${{F}_{M}}$ gets decreased.

2. A Charged Rod $P$ Attracts a Rod $R$ Whereas $P$ Repels another Charged Rod $Q$. What Type of Force is Developed Between  $Q$ and $R$ ?

Ans: Suppose that the rod $P$ is negatively charged. As it attracts rod$R$, it can be said that $R$ is positively charged. 

Also, since $P$ repels rod $Q$, it can be said that $Q$ is negatively charged. 

Clearly, the force between $Q$ and $R$ would be attractive in nature.

3. Which Physical Quantity has its S.I. Unit 

1. The S.I. unit of electric dipole moment is $Cm$. 

2. The S.I. unit of electric field intensity is $N/C$.

4. Define One Coulomb.

Ans: Charge on a body is said to be $1$ coulomb if it experiences a force of repulsion or attraction of $9\times {{10}^{9}}N$ from another equal charge when they are separated by a distance of $1m$.

Short Answer Questions in Electric Charges and Fields Important Questions with Answers PDF – 2 Marks

1. A Free Proton and a Free Electron are Placed in a Uniform Field. Which of the Two Experiences Greater Force and Greater Acceleration?

(Image will be Uploaded Soon)

Ans: Force on both the electron as well as the proton in the uniform field would be equal because $F=kq$ and it is known that charge on both electron and proton are the same. On the other hand, since acceleration is given by $a=\frac{F}{m}$ and as the mass of a proton is more than that of an electron, the acceleration of the electron would be more.

2. No Two Electric Lines of Force Can Intersect Each Other. Why?          

Ans:  Two electric lines of force can never intersect each other.  Suppose if they intersect, then at the point of intersection, there can be two tangents drawn.

These two tangents are supposed to represent two directions of electric field lines, which is not possible at a particular point.

3. The Graph Shows the Variation of Voltage $V$ Across the Plates of Two Capacitors $A$ and $B$ Versus Increase of Charge $Q$ Stored on Them. Which of the Two Capacitors Have Higher Capacitance? Give Reason for Your Answer.

Ans:  

It is known that $C=\frac{Q}{V}$

Clearly, for a given charge $Q$,

$C\propto \frac{1}{V}$

Now, from the given graph, it is seen that ${{V}_{A}}<{{V}_{B}}$.

Therefore, it can be concluded that ${{C}_{A}}>{{C}_{B}}$.

4. An Electric Dipole When Held at $30{}^\circ $ with Respect to a Uniform Electric Field of ${{10}^{4}}N/C$ Experiences a Torque of $9\times {{10}^{-26}}Nm$. Calculate Dipole Moment of the Dipole?

It is given that

τ = 9 x 10⁻²⁶Nm

Dipole moment $P$ needs to be calculated.

It is known that torque is given by \[\tau =PE\sin \theta \].

\[\Rightarrow P=\frac{\tau }{E\sin \theta }\]

\[\Rightarrow P=\frac{9\times {{10}^{-26}}}{{{10}^{4}}\times \sin 30{}^\circ }=\frac{9\times {{10}^{-26}}\times {{10}^{-4}}\times 2}{1}\]

\[\Rightarrow P=18\times {{10}^{-30}}Cm\]

a) Explain the Meaning of the Statement 'Electric Charge of a Body is Quantized’.

Ans:  The statement ‘electric charge of a body is quantized’ suggests that only integral \[(1,\text{ }2,3,4,...,n)\] number of electrons can be transferred from one body to another. 

This further suggests that charges are not transferred in fractions. 

Hence, a body possesses its total charge only in integral multiples of electric charges.

b) Why Can One Ignore the Quantization of Electric Charge When Dealing With Macroscopic I.e., Large Scale Charge?

Ans: When dealing with macroscopic or large-scale charges, the charges used are huge in number as compared to the magnitude of electric charge. 

Hence, the quantization of electric charge is of no use on a macroscopic scale. 

Therefore, it is ignored and considered that electric charge is continuous.

6. When a Glass Rod is Rubbed with a Silk Cloth, Charges Appear on Both. A Similar Phenomenon is Observed with Many Other Pairs of Bodies. Explain How This Observation is Consistent With the Law of Conservation of Charge

Ans: Rubbing is a phenomenon in which there is production of charges equal in magnitude but opposite in nature on the two bodies which are rubbed with each other. 

It is also seen that during such a phenomenon, charges are created in pairs. This phenomenon of charging is called as charging by friction. 

The net charge on a system of two rubbed bodies is equal to zero. This is because equal number of opposite charges in both the bodies annihilate each other. 

Clearly, when a glass rod is rubbed with a silk cloth, opposite natured charges appear on both these bodies. 

Thus, this phenomenon is consistent with the law of conservation of energy. As already mentioned, a similar phenomenon is observed with many other pairs of bodies too.

a) An Electric Field Line is a Continuous Curve. That is, a Field Line Cannot Have Sudden Breaks. Why Not?

Ans: An electrostatic field line is a continuous curve as it is known that a charge experiences a continuous force when traced in an electrostatic field. 

Also, the field line cannot have sudden breaks because the charge moves continuously and does not have the potential to jump from one point to another.

b) Explain Why Two Field Lines Never Cross Each Other at Any Point? 

Ans: Suppose two field lines cross each other at a particular point, then electric field intensity will show two directions at that point of intersection. 

This is impossible. Thus, two field lines can never cross each other.

8. An Electric Dipole with Dipole Moment $4\times {{10}^{-9}}Cm$ is Aligned at $30{}^\circ $ with Direction of a Uniform Electric Field of Magnitude $5\times {{10}^{4}}N{{C}^{-1}}$. Calculate the Magnitude of the Torque Acting on the Dipole.

It is given that:

Electric dipole moment, $p=4\times {{10}^{-9}}Cm$ 

Angle made by $p$ with uniform electric field, $\theta =30{}^\circ $ 

Electric field, $E=5\times {{10}^{4}}N{{C}^{-1}}$ 

Torque acting on the dipole is given by $\tau =pE\sin \theta $.

$\Rightarrow \tau =4\times {{10}^{-9}}\times 5\times {{10}^{4}}\times \sin 30{}^\circ $

$\Rightarrow \tau =20\times {{10}^{-5}}\times \frac{1}{2}$

$\Rightarrow \tau ={{10}^{-4}}Nm$ 

Thus, the magnitude of the torque acting on the dipole is ${{10}^{-4}}Nm$.

9. Figure Below Shows Tracks of Three Charged Particles in a Uniform Electrostatic Field. Give the Signs of the Three Charges. Which Particle Has the Highest Charge to Mass Ratio?

Ans: Since unlike charges attract and like charges repel each other, the particles 1 and 2 moving towards the positively charged plate are negatively charged whereas the particle 3 that moves towards the negatively charged plate is positively charged.

Since the charge to mass ratio is directly proportional to the amount of deflection for a given velocity, particle 3 would have the highest charge to mass ratio.

10. What is the Net Flux of the Uniform Electric Field of Exercise $1.15$ through a Cube of Side $20cm$ Oriented So That Its Faces Are Parallel to the Coordinate Planes?

Ans: It is given that all the faces of the cube are parallel to the coordinate planes. 

Clearly, the number of field lines entering the cube is equal to the number of field lines entering out of the cube. 

As a result, the net flux through the cube can be calculated to be zero.

11. Careful Measurement of the Electric Field at the Surface of a Black Box Indicate That the Net Outward Flux Through the Surface of the Box is $8.0\times {{10}^{3}}N{{m}^{2}}/C$.

a) What is the Net Charge Inside the Box?

It is given that,

The net outward flux through surface of the box, $\phi= 8.0\times{{10}^{3}}N{{m}^{2}}/C$.

For a body containing net charge $q$, flux is given by $\phi=\frac{q}{{{\varepsilon}_{0}}}$

${{\varepsilon }_{0}}=$ Permittivity of free space $=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$

Therefore, the charge $q$ is given by  $q=\phi {{\varepsilon }_{0}}$.

$\Rightarrow q=8.854\times {{10}^{-12}}\times 8.0\times {{10}^{3}}$

$\Rightarrow q=7.08\times {{10}^{-8}}$

$\Rightarrow q= 0.07\mu C$

Therefore, the net charge inside the box is $0.07\mu C$

b) If the Net Outward Flux Through the Surface of Box Were Zero, Could You Conclude That There Were No Charges Inside the Box? Why or Why Not?

Ans: No. The net flux entering out through a body depends on the net charge contained in the body. If the net flux is given to be zero, then it can be inferred that the net charge inside the body is zero. 

For the net charge associated with a body to be zero, the body can have equal amounts of positive and negative charges and thus, it is not necessary that there were no charges inside the box.

Short Answer Questions – 3 Marks

1. A Particle of Mass $m$ and Charge $q$ is Released from Rest in a Uniform Electric Field of Intensity $E$. Calculate the Kinetic Energy Attained by this Particle after Moving a Distance Between the Plates.                                                                                                            

We have the electrostatic force on a charge $q$ in electric field $E$ given by, 

$F=qE$.........................................(1)  

Also, we have Newton’s second law of motion given by, 

$F=ma$.........................................(2)

From (1) and (2), 

$a=\frac{qE}{m}$……………………………………………… (3)

We have the third equation of motion given by,

${{v}^{2}}-{{u}^{2}}=2as$

Since the charged particle is initially at rest,

$\Rightarrow {{v}^{2}}=2as$…………………………………… (4)

We have the expression for kinetic energy given by, 

$KE=\frac{1}{2}m{{v}^{2}}$……………………………………. (5)

Substituting (4) in (5) we get, 

$KE=\frac{1}{2}m\left( 2as \right)=mas$…………………………….. (6)

Substituting (3) in (6) to get, 

$KE=m\times \left( \frac{qE}{m} \right)\times s$

Therefore, we have the kinetic energy obtained by the particle of charge $q$ on moving a distance $s$ in electric field $E$ given by, 

$\therefore KE=qEs$

2. Two Charges $+q$ and $+9q$ are Separated by a Distance of $10a$. Find The Point on the Line Joining the Two Charges Where Electric Field is Zero. 

Let $P$ be the point (at $x$ distance from charge $+q$) on the line joining the given two charges where the electric field is zero. 

We know that the electric field at a point at $r$ distance from any charge $q$ is given by, 

$E=K\frac{q}{{{r}^{2}}}$

Electric field due to charge $+q$ at point $P$ would be, 

${{E}_{1}}=K\frac{\left( +q \right)}{{{x}^{2}}}$……………………………………………… (1)

Electric field due to charge $+9q$ at point $P$ would be, 

${{E}_{2}}=K\frac{\left( +9q \right)}{{{\left( 10a-x \right)}^{2}}}$………………………………………….. (2)

Since the net electric field at point $P$ is zero,

${{E}_{1}}+{{E}_{2}}=0$

$\Rightarrow \left| {{E}_{1}} \right|=\left| {{E}_{2}} \right|$

From (1) and (2),

$K\frac{\left( +q \right)}{{{x}^{2}}} = K\frac{\left( +9q \right)}{{{\left( 10a-x \right)}^{2}}}$

$\Rightarrow {{\left( 10a-x \right)}^{2}}=9{{x}^{2}}$

$\Rightarrow 10-x=3x$

$\Rightarrow 10a=4x$

$\therefore x=\frac{10}{4}a=2.5a$

Therefore, we found the point on the line joining the given two charges where the net electric field is zero to be at a distance $x=2.5a$ from charge $q$ and at a distance $10a-x=10a-2.5a=7.5a$ from charge $9q$.

a) Define the Term Dipole Moment $\vec{P}$ of an Electric Dipole Indicating its Direction and also Give Its S.i. Unit. 

Electric dipole moment is defined as the product of the magnitude of either of the two charges of the dipole and their distance of separation which would be the length of dipole. Mathematically,

$\vec{P}=2\vec{l}q$……………………………………………………. (1)

where $2\vec{l}$is the length of the dipole and $q$ is the charge. 

The direction of dipole is from $-ve$ to $+ve$ charge and its S.I. unit is coulomb meter$\left( Cm \right)$

b) An Electric Dipole is Placed in a Uniform Electric Field $\vec{E}$. Deduce the Expression for the Torque Acting on It.

Consider a dipole placed in uniform electric field $\vec{E}$ making an angle $\theta $ with it. 

Now, we know that the force acting on the given dipole will be the electrostatic force and this will be the cause for the resultant force. We have the expression for torque given by, 

$\tau =F\times x$…………………………………….. (2)

Where, $F$ is the force on the dipole and $x$ is the perpendicular distance. 

Where, force $F$ is given by, 

$F=qE$…………………………………………….. (3) 

From the figure we have, 

$\sin \theta =\frac{BN}{AB}$

$\Rightarrow BN=AB\sin \theta =2l\sin \theta $

But $BN$here is the perpendicular distance $x$, so, equation (2) becomes,

$\tau =qE\times 2l\sin \theta =\left( 2lq \right)E\sin \theta $

But from (1), $P=2lq$

Now, we could give the torque on the dipole as, 

$\therefore \tau =PE\sin \theta =\vec{P}\times \vec{E}$

4. A Sphere ${{S}_{1}}$ of Radius ${{R}_{1}}$ Encloses a Charge $Q$. If there is another Concentric Sphere${{S}_{2}}$ of Radius ${{R}_{2}}\left( {{R}_{2}}>{{R}_{1}} \right)$  and there is no Additional Change Between ${{S}_{1}}$ and ${{S}_{2}}$, then Find the Ratio of Electric Flux through ${{S}_{1}}$  and ${{S}_{2}}$.

We may recall that the expression for electric flux through a surface enclosing charge $q$ by Gauss’s law is given by, 

$\phi =\frac{q}{{{\varepsilon }_{0}}}$

Where, ${{\varepsilon }_{0}}$is the permittivity of the medium. 

Now the electric flux through sphere ${{S}_{1}}$ is given by, 

${{\phi }_{{{S}_{1}}}}=\frac{Q}{{{\varepsilon }_{0}}}$ …………………………………….. (1)

Since there is no additional charge between the given two spheres, the flux through sphere ${{S}_{2}}$ is given by, 

${{\phi }_{{{S}_{2}}}}=\frac{Q}{{{\varepsilon }_{0}}}$…………………………………….. (2)

We could now get the ratio of flux through spheres ${{S}_{1}}$ and ${{S}_{2}}$,

$\frac{{{\phi }_{{{S}_{1}}}}}{{{\phi }_{{{S}_{2}}}}}=\frac{\frac{Q}{{{\varepsilon }_{0}}}}{\frac{Q}{{{\varepsilon }_{0}}}}$

$\therefore \frac{{{\phi }_{{{S}_{1}}}}}{{{\phi }_{{{S}_{2}}}}}=\frac{1}{1}$

Therefore, we find the required ratio to be $1:1$ .

5. Electric Charge is Uniformly Distributed on the Surface of a Spherical Balloon. Show How Electric Intensity and Electric Potential Vary 

a) on the surface 

Electric field intensity on the surface of the balloon would be,

$E=\frac{\sigma }{{{\varepsilon }_{0}}}$ 

Electric potential on the surface of the balloon would be, 

$\text{ }V=\frac{Kq}{R}$

b) inside 

Electric field intensity inside the balloon would be,

$E=0\text{  }$

Electric potential inside the balloon would be, 

$V=\frac{Kq}{R}$

c) outside.

Electric field intensity outside the balloon would be,

$E=\frac{\sigma }{{{\varepsilon }_{0}}}\frac{{{R}^{2}}}{{{r}^{2}}}\text{  }$

Electric potential outside the balloon would be

$V=\frac{Kq}{r}$

We could represent this variation graphically as, 

For electric field,  

For electric potential, 

6. Two Point Electric Charges $q$ and $2q$ are Kept at a Distance $d$ Apart from Each Other in Air. a Third Charge $Q$ is to be Kept Along the Same Line in Such a Way That the Net Force Acting On $q$ and $2q$ is Zero. Calculate the Position of Charge $Q$ in Terms of $q$ and $d$.

For the net force on charge $q$ and $2q$ to be zero, the third charge should be negative since other two given charges are positive.  

The force between two charges ${{q}_{1}}$ and ${{q}_{2}}$ separated by a distance$r$ is given by Coulomb’s law as, 

$F=K\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$

Force acting on charge $Q$ due to $q$$=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{qQ}{{{x}^{2}}}$

Force acting on charge $Q$due to $2q$$=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2qQ}{{{\left( d-x \right)}^{2}}}$

Now for the given system to be in equilibrium,

$K\frac{Qq}{{{x}^{2}}}=K\frac{2Qq}{{{\left( d-x \right)}^{2}}}$……………………………………………….. (1)

From equations (1) and (2) we get, 

$\frac{1}{{{x}^{2}}}=\frac{2}{{{\left( d-x \right)}^{2}}}$

$\Rightarrow 2{{x}^{2}}={{\left( d-x \right)}^{2}}$

$\Rightarrow \sqrt{2}x=d-x$

$\therefore x=\frac{d}{\sqrt{2}+1}$

So, we found that the new charge $Q$ should be kept between the given two charges at a distance of $x=\frac{d}{\sqrt{2}+1}$ from charge $q$.

7. What is the Force Between Two Small Charged Spheres Having Charges of $2\times {{10}^{-7}}C$ and $3\times {{10}^{-7}}C$ placed $30cm$ Apart in Air? 

We are given:

Charge of the first sphere, ${{q}_{1}}=2\times {{10}^{-7}}C$

Charge of the second sphere, ${{q}_{2}}=3\times {{10}^{-7}}C$

Distance between the two spheres, $r=30cm=0.3m$

Electrostatic force between the spheres is given by Coulomb’s law as,

$F=\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$ ……………………………… (1)

Where, ${{\varepsilon }_{0}}=$Permittivity of free space and,$\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{2}}$

Substituting the given values in (1), we get, 

$F=\frac{9\times {{10}^{9}}\times 2\times {{10}^{-7}}\times 3\times {{10}^{-7}}}{{{\left( 0.3 \right)}^{2}}}$

$\therefore F=6\times {{10}^{-3}}N$

Hence, force between the two small charged spheres is found to have a magnitude of$6\times {{10}^{-3}}N$ . 

Since both the given charges are positive, the resultant force would be repulsive as like charges repel each other. 

8. The Electrostatic Force on a Small Sphere of Charge $0.4\mu C$ Due to another Small Sphere of Charge $-0.8\mu C$ in Air is $0.2N$.

a) What is the Distance Between the Two Spheres? 

Electrostatic force on the first sphere,$F=0.2N$ 

Charge on the first sphere, ${{q}_{1}}=0.4\mu C=0.4\times {{10}^{-6}}C$

Charge on the second sphere,${{q}_{2}}=-0.8\mu C=-0.8\times {{10}^{-6}}C$ 

Electrostatic force between the spheres could be given by Coulomb’s law as,

Where, ${{\varepsilon }_{0}}=$Permittivity of free space and,$\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{2}}$ 

Rearranging (1) we get, 

$\Rightarrow {{r}^{2}}=\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{0}}F}$

Substituting the given values, 

$\Rightarrow {{r}^{2}}=\frac{0.4\times {{10}^{-6}}\times -0.8\times {{10}^{-6}}\times 9\times {{10}^{9}}}{0.2}$

$\Rightarrow {{r}^{2}}=144\times {{10}^{-4}}$

$\Rightarrow r=\sqrt{144\times {{10}^{-4}}}$

$\therefore r=0.12m$

Therefore, we found the distance between the given two spheres to be $0.2m$.

b) What is the Force on the Second Sphere Due to the First?

Ans: Since, both the spheres attract each other with the same force, the force on the second sphere due to the first would be $0.2N$. 

9. A Polythene Piece Rubbed With Wool is Found to Have a Negative Charge Of $3\times {{10}^{-7}}C$.

a) Estimate the Number of Electrons Transferred (from Which to Which?)

When polythene is rubbed against wool, certain number of electrons get transferred from wool to polythene. 

Hence, wool becomes positively charged on loosing electrons and polythene becomes negatively charged on gaining them.

Charge on the polythene piece, 

$q=-3\times {{10}^{-7}}C$ 

Charge of an electron, 

$e=-1.6\times {{10}^{-19}}C$ 

Let the number of electrons transferred from wool to polythene be $n$, then, from the property of quantization of charge we have, 

$q=ne$ 

$\Rightarrow n=\frac{q}{e}$

Now, on substituting the given values, we get, 

$\Rightarrow n=\frac{-3\times {{10}^{-7}}}{-1.6\times {{10}^{-19}}}$

$\therefore n=1.87\times {{10}^{12}}$

Therefore, the number of electrons transferred from wool to polythene is found to be$1.87\times {{10}^{12}}$ .

b) Is There a Transfer of Mass from Wool to Polythene?

Ans: Yes, during the transfer of electrons from wool to polythene, along with charge, mass is also transferred. 

Let $m$ be the mass being transferred in the given case and ${{m}_{e}}$ be the mass of the electron, then,

$m={{m}_{e}}\times n$ 

$\Rightarrow m=9.1\times {{10}^{-31}}\times 1.85\times {{10}^{12}}$

$\therefore m=1.706\times {{10}^{-18}}kg$

Hence, we found that a negligible amount of mass does get transferred from wool to polythene.

10. Consider a Uniform Electric Field $E=3\times {{10}^{3}}\hat{i}N/C$. 

a) What is the Flux of This Field Through a Square of Side $10cm$ Whose Plane is Parallel to the Y-Z Plane? 

Electric field intensity,$\vec{E}=3\times {{10}^{3}}\hat{i}N/C$ 

Magnitude of electric field intensity,$\left| {\vec{E}} \right|=3\times {{10}^{3}}N/C$ 

Side of the square,$a=10cm=0.1m$ 

Area of the square, $A={{a}^{2}}=0.01{{m}^{2}}$ 

Since the plane of the square is parallel to the y-z plane, the normal to its plane would be directed in the x direction. So, angle between normal to the plane and the electric field would be,$\theta =0{}^\circ $ 

We know that the flux through a surface is given by the relation, 

$\phi =EA\cos \theta $

Substituting the given values, we get, 

$\Rightarrow \phi =3\times {{10}^{3}}\times 0.01\times \cos 0{}^\circ $

$\therefore \phi =30N{{m}^{2}}/C$

Now, we found the net flux through the given surface to be, $\phi =30N{{m}^{2}}/C$.

b) What is the Flux Through the Same Square if the Normal to Its Plane Makes $60{}^\circ $angle with the x-axis? 

When the plane makes an angle of $60{}^\circ $ with the x-axis, the flux through the given surface would be,

$\Rightarrow \phi =3\times {{10}^{3}}\times 0.01\times \cos 60{}^\circ $

$\Rightarrow \phi =30\times \frac{1}{2}$

$\therefore \phi =15N{{m}^{2}}/C$

So, we found the flux in this case to be,$\phi =15N{{m}^{2}}/C$.  

11. A Point Charge $+10\mu C$  is a Distance $5cm$ Directly Above the Center of a Square of Side $10cm$, as Shown in Fig. 1.34. What is the Magnitude of the Electric Flux Through the Square? (Hint: Think of the Square as One Face of a Cube With Edge $10cm$) 

Considering square as one face of a cube of edge $10cm$ with a charge $q$ at its center, according to Gauss's theorem for a cube, total electric flux is through all its six faces.

${{\phi }_{total}}=\frac{q}{{{\varepsilon }_{0}}}$

The electric flux through one face of the cube could be given by, $\phi =\frac{{{\phi }_{total}}}{6}$

$\Rightarrow \phi =\frac{1}{6}\frac{q}{{{\varepsilon }_{0}}}$ 

Permittivity of free space, ${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$.

The net charge enclosed, $q=10\mu C=10\times {{10}^{-6}}C$.

Substituting the values given in the question, we get, 

$\phi =\frac{1}{6}\times \frac{10\times {{10}^{-6}}}{8.854\times {{10}^{-12}}}$

$\therefore \phi =1.88\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}$

Therefore, electric flux through the square is found to be $1.88\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}$.

12. A Point Charge of $2.0\mu C$ is Kept at the Center of a Cubic Gaussian Surface of Edge Length $9cm$. What is the Net Electric Flux Through the Surface? 

Let us consider one of the faces of the cubical Gaussian surface considered, which would be a square. 

Since, a cube has six such square faces in total, we could say that the flux through one surface would be one-sixth the total flux through the gaussian surface considered. 

The net flux through the cubical Gaussian surface by Gauss’s law is given by, 

So, the electric flux through one face of the cube would be, 

\[\phi =\frac{{{\phi }_{total}}}{6}\]

$\Rightarrow \phi =\frac{1}{6}\frac{q}{{{\varepsilon }_{0}}}$……………………………….. (1)

But we have, permittivity of free space, ${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$. 

Charge enclosed, $q=10\mu C=10\times {{10}^{-6}}C$.

Substituting the given values in (1) we get, 

Therefore, electric flux through the square surface is $1.88\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}$.

13. A Point Charge Causes an Electric Flux of $-1.0\times {{10}^{3}}N{{m}^{2}}/C$ to Pass Through a Spherical Gaussian Surface Of $10cm$ Radius Centered on the Charge. 

a) If the Radius of the Gaussian Surface Were Doubled, How Much Flux Would Pass Through the Surface? 

Electric flux due to the given point charge, 

$\phi =-1.0\times {{10}^{3}}N{{m}^{2}}/C$ 

Radius of the Gaussian surface enclosing the point charge,

$r=10.0cm$ 

Electric flux piercing out through a surface depends on the net charge enclosed by the surface from Gauss’s law.

It is independent of the dimensions of the arbitrary surface assumed to enclose this charge. 

If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e.,$-{{10}^{3}}N{{m}^{2}}/C$ .

b) What is the Value of the Point Charge?

Ans:  

Electric flux is given by the relation,

$q=$ net charge enclosed by the spherical surface 

Permittivity of free space, ${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$ 

$\Rightarrow q=\phi {{\varepsilon }_{0}}$

Substituting the given values,

$\Rightarrow q=-1.0\times {{10}^{3}}\times 8.854\times {{10}^{-12}}=-8.854\times {{10}^{-9}}C$

$\therefore q=-8.854nC$

Therefore, the value of the point charge is found to be $-8.854nC$.

14. A Conducting Sphere of Radius $10cm$has an Unknown Charge. If the Electric Field at a Point $20cm$ from the Center of the Sphere of Magnitude $1.5\times {{10}^{3}}N/C$ is Directed Radially Inward, What Is the Net Charge on the Sphere?

We have the relation for electric field intensity$E$ at a distance $d$ from the center of a sphere containing net charge $q$is given by,

$E=\frac{q}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}$ ……………………………………………… (1)

Where, 

Net charge, 

$q=1.5\times {{10}^{3}}N/C$

Distance from the center, 

$d=20cm=0.2m$ 

Permittivity of free space,

${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$ 

$\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$

From (1), the unknown charge would be, 

$q=E\left( 4\pi {{\varepsilon }_{0}} \right){{d}^{2}}$

Substituting the given values we get, 

$q=\frac{1.5\times {{10}^{3}}\times {{\left( 0.2 \right)}^{2}}}{9\times {{10}^{9}}}=6.67\times {{10}^{-9}}C$

$\therefore q=6.67nC$

Therefore, the net charge on the sphere is found to be $6.67nC$.

15. A Uniformly Charged Conducting Sphere of $2.4m$ Diameter Has a Surface Charge Density of $80.0\mu C/{{m}^{2}}$ .

a) Find the Charge on the Sphere. 

Diameter of the sphere, 

$d=2.4m$ 

Radius of the sphere,

$r=1.2m$ 

Surface charge density, 

\[\sigma =80.0\mu C/{{m}^{2}}=80\times {{10}^{-6}}C/{{m}^{2}}\] 

Total charge on the surface of the sphere,

$Q=\text{Charge density }\times \text{ Surface area}$ 

$\Rightarrow \text{Q}=\sigma \times \text{4}\pi {{\text{r}}^{2}}=80\times {{10}^{-6}}\times 4\times 3.14\times {{\left( 1.2 \right)}^{2}}$

$\therefore Q=1.447\times {{10}^{-3}}C$

Therefore, the charge on the sphere is found to be $1.447\times {{10}^{-3}}C$.

b) What is the Total Electric Flux Leaving the Surface of the Sphere?

Total electric flux $\left( {{\phi }_{total}} \right)$ leaving out the surface containing net charge $Q$ is given by Gauss’s law as, 

${{\phi }_{total}}=\frac{Q}{{{\varepsilon }_{0}}}$…………………………………………………. (1)

Where, permittivity of free space,

${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$

We found the charge on the sphere to be, 

$Q=1.447\times {{10}^{-3}}C$

Substituting these in (1), we get, 

${{\phi }_{total}}=\frac{1.447\times {{10}^{-3}}}{8.854\times {{10}^{-12}}}$

$\therefore {{\phi }_{total}}=1.63\times {{10}^{-8}}N{{C}^{-1}}{{m}^{2}}$

Therefore, the total electric flux leaving the surface of the sphere is found to be $1.63\times {{10}^{-8}}N{{C}^{-1}}{{m}^{2}}$.

16. An Infinite Line Charge Produces a Field of Magnitude $9\times {{10}^{4}}N/C$ at a Distance of $2cm$. Calculate the Linear Charge Density.

Electric field produced by the given infinite line charge at a distance $d$having linear charge density$\lambda $ could be given by the relation,

$E=\frac{\lambda }{2\pi {{\varepsilon }_{0}}d}$ 

$\Rightarrow \lambda =2\pi {{\varepsilon }_{0}}Ed$…………………………………….. (1)

$d=2cm=0.02m$  

$E=9\times {{10}^{4}}N/C$ 

Substituting these values in (1) we get, 

$\lambda =2\pi \left( 8.854\times {{10}^{-12}} \right)\left( 9\times {{10}^{4}} \right)\left( 0.02 \right)$

$\therefore \lambda =10\times {{10}^{-8}}C/m$

Therefore, we found the linear charge density to be $10\times {{10}^{-8}}C/m$.

17. Which Among the Curves Shown in Fig. 1.35 Cannot Possibly Represent Electrostatic Field Lines?

The field lines showed in (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor which is a characterizing property of electric field lines.

  (Image will be Uploaded Soon)

The lines showed in (b) do not represent electrostatic field lines because field lines cannot emerge from a negative charge and cannot terminate at a positive charge since the direction of the electric field is from positive to negative charge. 

The field lines showed in (c) do represent electrostatic field lines as they are directed outwards from positive charge in accordance with the property of electric field. 

The field lines showed in (d) do not represent electrostatic field lines because electric field lines should not intersect each other.

The field lines showed in (e) do not represent electrostatic field lines because electric field lines do not form closed loops

18. Suppose that the Particle in Exercise in 1.33 Is an Electron Projected With Velocity ${{v}_{x}}=2.0\times {{10}^{6}}m{{s}^{-1}}$. If $E$ Between the Plates Separated by $0.5cm$ is $9.1\times {{10}^{2}}N/C$, Where Will the Electron Strike the Upper Plate? ($\left| e \right|=1.6\times {{10}^{-19}}C,{{m}_{e}}=9.1\times {{10}^{-31}}kg$ )

We are given the velocity of the particle, ${{v}_{x}}=2.0\times {{10}^{6}}m{{s}^{-1}}$

Separation between the two plates, $d=0.5cm=0.005m$ 

Electric field between the two plates, $E=9.1\times {{10}^{2}}N/C$ 

Charge on an electron, $e=1.6\times {{10}^{-19}}C$ 

mass of an electron, ${{m}_{e}}=9.1\times {{10}^{-31}}kg$ 

Let$s$be the deflection when the electron strikes the upper plate at the end of the plate $L$, then, we have the deflection given by, 

$s=\frac{qE{{L}^{2}}}{2m{{v}_{x}}}$ 

$\Rightarrow L=\sqrt{\frac{2dm{{v}_{x}}}{qE}}$

$L=\sqrt{\frac{2\times 0.005\times 9.1\times {{10}^{-31}}\times {{\left( 2.0\times {{10}^{6}} \right)}^{2}}}{1.6\times {{10}^{-19}}\times 9.1\times {{10}^{2}}}}=\sqrt{0.025\times {{10}^{-2}}}=\sqrt{2.5\times {{10}^{-4}}}$

$\therefore L=1.6\times {{10}^{-2}}=1.6cm$

Therefore, we found that the electron will strike the upper plate after travelling a distance of $1.6cm$.

Short Answer Questions – 5 Marks

a) The Expression of Electric Field $\vec{E}$due to a Point Charge at Any Point Near to it Is Defined By \[\vec{E}=\underset{q\to 0}{\mathop{\lim }}\,\frac{{\vec{F}}}{q}\] Where $q$  is the Test Charge and $\vec{F}$ is the force acting on it. What is the Significance of $\underset{q\to 0}{\mathop{\lim }}\,$ in this Expression?

Ans: The significance of  $\underset{q\to 0}{\mathop{\lim }}\,$ is that the test charge should be vanishingly small so that it is not disrupting the presence of the source charge.

b) Two Charges Each of Magnitude $2\times {{10}^{-7}}C$ but Opposite in Sign Forms a System. These Charges Are Located at Points \[A\text{ }\left( 0,0,-10 \right)\] and \[B\text{ }\left( 0,0,+10 \right)\] Respectively. Distances Are Given in Cm. What Is the Total Charge and Electric Dipole Moment of the System?

Ans.: 

Total charge of the system $=(+2\times {{10}^{-7}})+(-2\times {{10}^{-7}})=0$.

Electric dipole moment is:

$P=q\times 2l$

\[P=2\times {{10}^{-7}}\times 20\times {{10}^{-2}}\]

$P=4\times {{10}^{-8}}cm$

Also, the direction of electric dipole moment is along the negative z-axis.

A) Sketch Electric Lines of Force Due To 

i) Isolated Positive Charge (i.e., \[q>0\]) and 

ii) Isolated Negative Charge (i.e., $q<0$).

The sketch of isolated positive charge and isolated negative charge are as follows:

b) Two-point Charges $q$ and $-q$ are Placed at a Distance of \[2a\] apart. Calculate the Electric Field at a Point $P$ Situated at a Distance $r$ Along the Perpendicular Bisector of the Line Joining the Charges. What Is the Electric Field When \[r\text{ }>>\text{ }a\]?

As we know,

$|{{\vec{E}}_{+q}}|=\frac{kq}{{{r}^{2}}+{{a}^{2}}}$

$|{{\vec{E}}_{-q}}|=\frac{kq}{{{r}^{2}}+{{a}^{2}}}$

Since, $|{{\vec{E}}_{+q}}|=|{{\vec{E}}_{-q}}|$

$|{{\vec{E}}_{net}}|=\sqrt{\mathop{{{E}_{+q}}}^{2}+\mathop{{{E}_{-q}}}^{2}+2{{E}_{+q}}{{E}_{-q}}\cos 2\theta }$

$|{{\vec{E}}_{net}}|=\sqrt{\mathop{2{{E}_{+q}}}^{2}+\mathop{2{{E}_{+q}}}^{2}\cos 2\theta }$

$|{{\vec{E}}_{net}}|=\sqrt{\mathop{2{{E}_{+q}}}^{2}\left( 1+\cos 2\theta  \right)}$

$|{{\vec{E}}_{net}}|=\sqrt{\mathop{2{{E}_{+q}}}^{2}\left( 2{{\cos }^{2}}\theta  \right)}$

$|{{\vec{E}}_{net}}|=\sqrt{\mathop{4{{E}_{+q}}}^{2}\left( {{\cos }^{2}}\theta  \right)}$

\[|{{\vec{E}}_{net}}|=2{{E}_{+q}}\cos \theta \]

$|{{\vec{E}}_{net}}|=2{{E}_{+q}}\frac{a}{\sqrt{{{r}^{2}}+{{a}^{2}}}}$

$|{{\vec{E}}_{net}}|=2\frac{kq}{\sqrt{{{r}^{2}}+{{a}^{2}}}}\frac{a}{\sqrt{{{r}^{2}}+{{a}^{2}}}}$

$|{{\vec{E}}_{net}}|=\frac{2akq}{{{\left( {{r}^{2}}+{{a}^{2}} \right)}^{\frac{3}{2}}}}$

$|{{\vec{E}}_{net}}|=\frac{k\vec{P}}{{{\left( {{r}^{2}}+{{a}^{2}} \right)}^{\frac{3}{2}}}}$

For, $r>>a$ ($a$ can be neglected)

Therefore, we get,

$|{{\vec{E}}_{net}}|=\frac{k\vec{P}}{{{r}^{3}}}$

a) What Is an Equi-Potential Surface? Show That the Electric Field Is Always Directed Perpendicular to an Equi-Potential Surface.

An equipotential surface is a surface that has the same potential throughout.

\[dW=F\cdot dx\]

$dW=\left( q\cdot E \right)\cdot dx$

(Force on the test charge, $F=\left( q\cdot E \right)$)

Since work done is moving a test charge along an equipotential surface is always zero,

$0=\left( q\cdot E \right)\cdot dx$

\[E\cdot dx=0\]

\[\Rightarrow E\bot dx\]

b) Derive an Expression for the Potential at a Point Along the Axial Line of a Short Electric Dipole

Consider an electric dipole of dipole length \[2a\] and point \[P\] on the axial line such that $OP=r$, where \[O\] is the centre of the dipole.

Electric potential at point $P$ due to the dipole is given by:

$V={{V}_{PA}}+{{V}_{PB}}$

$V=\frac{K(-q)}{(r+a)}+\frac{K(+q)}{(r-a)}$

$V=Kq\left[ \frac{1}{r-a}-\frac{1}{r+a} \right]$

$V=Kq\left[ \frac{r+a-r+a}{(r-a)(r+a)} \right]$

$V=Kq\frac{2a}{{{r}^{2}}-{{a}^{2}}}$       ($\because P=2aq$)

$V=\frac{KP}{{{r}^{2}}-{{a}^{2}}}$

For a dipole having short length, $a$ can be neglected.

This gives,

$V=\frac{KP}{\mathop{r}^{2}}$

4. Check if the Ratio \[\frac{K{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}\] is Dimensionless. Look up at the Table of Physical Constants and Determine the Value of This Ratio. What Does This Ratio Signify?

The given ratio is \[\frac{K{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}\].

Where, $G$ is Gravitational constant. Its unit is $N{{m}^{2}}k{{g}^{-2}}$.

${{m}_{e}}$ and ${{m}_{p}}$ are the masses of electron and proton respectively. Their unit is $kg$.

$e$ is the electric charge. Its unit is $C$.

${{\varepsilon }_{o}}$ is the permittivity of free space. Its unit is $N{{m}^{2}}{{C}^{-2}}$.

Therefore, the unit of the given ratio \[\frac{K{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}\] is

$=\frac{[N{{m}^{2}}{{C}^{-2}}][{{C}^{2}}]}{[N{{m}^{2}}k{{g}^{-2}}][kg][kg]}$

And its dimensions can be related to $=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]$

Hence, the given ratio is dimensionless.

$e=1.6\times {{10}^{-19}}C$

$G=6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$

${{m}_{e}}=9.1\times {{10}^{-31}}kg$

${{m}_{p}}=1.66\times {{10}^{-27}}kg$

Hence, the numerical value of the given ratio is

\[\frac{K{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}=\frac{9\times {{10}^{9}}\times {{(1.6\times {{10}^{-19}})}^{2}}}{6.67\times {{10}^{-11}}\times 9.1\times {{10}^{-31}}\times 1.66\times {{10}^{-27}}}\approx 2.3\times {{10}^{39}}\]

This ratio is showing the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.

5. Four-point Charges $\mathop{q}_{A}=2\mu C$, $\mathop{q}_{B}=-5\mu C$, $\mathop{q}_{C}=2\mu C$, $\mathop{q}_{D}=-5\mu C$ are Located at the Corners of a Square ABCD of Side \[10\] cm. What is the Force on a Charge of \[1\mu C\] Placed at the Centre of the Square?

Ans.  

In the given figure, there is a square having length of each side is \[10cm\] and four charges placed at its corners. O is the centre of the square.

AB, BC, CD and AD are the sides of the square. Each of length is \[10cm\]

AC and BD are the diagonals of the square of length $10\sqrt{2}cm$.

AO, OB, OC, OD are of length $5\sqrt{2}cm$.

A charge of amount \[1\mu C\] is placed at the centre of square. 

There is repulsion force between charges located at A and O is equal in magnitude but having opposite direction relative to the repulsion force between the charges located at C and O. Hence, they will cancel each other forces. 

Similarly, there is attraction force between charges located at B and O equal in magnitude but having opposite direction relative to the attraction force between the charge placed at D and O. Hence, they also cancel each other forces. 

Therefore, the net force due to the four charges placed at the corners of the square on \[1\mu C\] charge which is placed at centre O is zero.

a) Two-point Charges ${{q}_{A}}=3\mu C$ and $\mathop{q}_{B}=-3\mu C$ are Located $20cm$ Apart in Vacuum. What Is the Electric Field at the Midpoint O of the Line Ab Joining the Two Charges? 

O is the mid-point of line AB. Distance between the two charges i.e., $AB=20cm$

Therefore, $OA=OB=10cm$.

Electric field at point O due to $+3\mu C$ charge:

\[{{E}_{1}}=\frac{3\times {{10}^{-6}}}{4\pi {{\varepsilon }_{o}}\mathop{(AO)}^{2}}\]

\[{{E}_{1}}=\frac{3\times {{10}^{-6}}}{4\pi {{\varepsilon }_{o}}\mathop{(10\times {{10}^{-2}})}^{2}}N{{C}^{-1}}\] along OB.

Where, ${{\varepsilon }_{o}}$ is the permittivity of free space.

$\frac{1}{4\pi {{\varepsilon }_{o}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$.

Electric field at point O due to $-3\mu C$ charge:

\[{{E}_{2}}=\frac{3\times {{10}^{-6}}}{4\pi {{\varepsilon }_{o}}\mathop{(OB)}^{2}}\]

${{E}_{2}}=\frac{3\times {{10}^{-6}}}{4\pi {{\varepsilon }_{o}}\mathop{(10\times {{10}^{-2}})}^{2}}N{{C}^{-1}}$ along OB.

\[\therefore E={{E}_{1}}+{{E}_{2}}\]

$E=2\times \frac{9\times {{10}^{9}}\times 3\times {{10}^{-6}}}{\mathop{(10\times {{10}^{-2}})}^{2}}N{{C}^{-1}}$ 

[since, ${{E}_{1}}$ and ${{E}_{2}}$ having same values, so, the value is multiplied with $2$]

$E=5.4\times {{10}^{6}}N{{C}^{-1}}$ along OB.

Therefore, the electric field at mid-point O is \[5.4\times {{10}^{6}}N{{C}^{-1}}\] along OB.

b) If a Negative Test Charge of Magnitude \[1.5\times {{10}^{-9}}C\] is Placed at This Point, What is the Force Experienced by the Test Charge?

A test charge \[1.5\times {{10}^{-9}}C\] is placed at mid-point O.

$q=1.5\times {{10}^{-9}}C$

Force experienced by test charge, $F=qE$

\[F=1.5\times {{10}^{-9}}\times 5.4\times {{10}^{6}}\]

\[F=8.1\times {{10}^{-3}}N\]

The force is aimed along line OA. The negative test charge is repelled by the charge located at point B but attracted towards A. 

Therefore, the force felt by the test charge is \[8.1\times {{10}^{-3}}N\] along OA.

7. A System Has Two Charges \[{{q}_{A}}=2.5\times {{10}^{-7}}C\] and \[{{q}_{B}}=-2.5\times {{10}^{-7}}C\] located at points \[A\text{ }\left( 0,0,-15 \right)\] and \[\text{B }\left( 0,0,15 \right)\] Respectively. What Are the Total Charge and Electric Dipole Moment of the System?

Two charges are located at their respective position.

The value of charge at A, \[{{q}_{A}}=2.5\times {{10}^{-7}}C\]

The value of charge at B, \[{{q}_{B}}=-2.5\times {{10}^{-7}}C\]

Net amount of charge, ${{q}_{net}}={{q}_{A}}+{{q}_{B}}$

${{q}_{net}}=+2.5\times {{10}^{-7}}-2.5\times {{10}^{-7}}$

${{q}_{net}}=0$

The distance between two charges at A and B,

\[d=15+15=30cm\]

The electric dipole moment of the system is given by

$P=\mathop{q}_{A}\times d=\mathop{q}_{B}\times d$

$P=2.5\times {{10}^{-7}}\times 0.3$

$P=7.5\times {{10}^{-8}}Cm$ along z-axis.

Therefore, $7.5\times {{10}^{-8}}Cm$ is the electric dipole moment of the system and it is along positive z-axis.

a) Two Insulated Charged Copper Spheres a and B Have Their Centres Separated by a Distance of $50cm$. What Is the Mutual Force of Electrostatic Repulsion If the Charge on Each is $6.5\times {{10}^{-7}}C$? The Radii of A and B Are Negligible Compared to the Distance of Separation. 

Charges on both A and B is equal to ${{q}_{A}}={{q}_{B}}=6.5\times {{10}^{-7}}C$

Distance between the centres of the spheres is given as $r=50cm=0.5m$

It is known that the force of repulsion between the two spheres would be

$F=\frac{{{q}_{A}}{{q}_{B}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$

${{\varepsilon }_{o}}$ is the permittivity of the free space

Substituting the known values in the above expression, 

$F=\frac{9\times {{10}^{9}}\times {{(6.5\times {{10}^{-7}})}^{2}}}{{{(0.5)}^{2}}}=1.52\times {{10}^{-2}}N$

The mutual force of electrostatic repulsion between the two spheres is $1.52\times {{10}^{-2}}N$.

b)What is the Force of Repulsion if Each Sphere is Charged Double the Above Amount, and the Distance Between Them is Halved?

Next, it is told that the charges on both the spheres are doubled and the distance between the centres of the spheres is halved. Thus,

${{q}_{A}}'={{q}_{B}}'=2\times 6.5\times {{10}^{-7}}=13\times {{10}^{-7}}C$

$r'=\frac{1}{2}(0.5)=0.25m$

Now, substituting this in the relation for force, 

$F'=\frac{{{q}_{A}}'{{q}_{B}}'}{4\pi {{\varepsilon }_{0}}r{{'}^{2}}}=\frac{9\times {{10}^{9}}\times {{(13\times {{10}^{-7}})}^{2}}}{{{(0.25)}^{2}}}=0.243N$

The new mutual force of electrostatic repulsion between the two spheres is $0.243N$.

9. Suppose the Spheres A and B in Exercise $1.12$ Have Identical Sizes. a Third Sphere of the Same Size but Uncharged Is Brought in Contact With the First, Then Brought in Contact With the Second, and Finally Removed from Both. What Is the New Force of Repulsion Between A and B?

Distance between the spheres A and B is $r=0.5m$

The charge on each sphere initially is ${{q}_{A}}={{q}_{B}}=6.5\times {{10}^{-7}}C$

Now, when uncharged sphere C is made to touch the sphere A, the amount of charge from A will get transferred to the sphere C, making both A and C to have equal charges in them. Clearly,

${{q}_{A}}'={{q}_{C}}=\frac{1}{2}(6.5\times {{10}^{-7}})=3.25\times {{10}^{-7}}C$

Now, when the sphere C is made to touch the sphere B, there is similar transfer of charge making both C and B to have equal charges in them. Clearly,

${{q}_{C}}'={{q}_{B}}'=\frac{3.25\times {{10}^{-7}}+6.5\times {{10}^{-7}}}{2}=4.875\times {{10}^{-7}}C$

Thus, the new force of repulsion between the spheres A and B will turn out to be

$F'=\frac{{{q}_{A}}'{{q}_{B}}'}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}=\frac{9\times {{10}^{9}}\times 3.25\times {{10}^{-7}}\times 4.875\times {{10}^{-7}}}{{{(0.5)}^{2}}}=5.703\times {{10}^{-3}}N$

10. Two Large, Thin Metal Plates Are Parallel and Close to Each Other. on Their Inner Faces, the Plates Have Surface Charge Densities of Opposite Signs and of Magnitude $17.0\times {{10}^{-22}}C{{m}^{-2}}$. What is $E$ in the Outer Region of the First Plate? What is $E$ in the Outer Region of the Second Plate? What is E Between the Plates?

The given nature of metal plates is represented in the figure below: 

Here, A and B are two parallel plates kept close to each other. The outer region of plate A is denoted as \[I\], outer region of plate B is denoted as \[III\], and the region between the plates, A and B, is denoted as \[II\].

Charge density of plate A, $\sigma =17.0\times {{10}^{-22}}C/{{m}^{2}}$ 

Charge density of plate B, $\sigma =-17.0\times {{10}^{-22}}C/{{m}^{2}}$

In the regions \[I\] and \[III\], electric field E is zero. This is because the charge is not enclosed within the respective plates.

Now, the electric field $E$ in the region \[II\] is given by

$E=\frac{|\sigma |}{{{\varepsilon }_{0}}}$ 

where, 

${{\varepsilon }_{0}}=$ Permittivity of free space $=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$ 

$E=\frac{17.0\times {{10}^{-22}}}{8.854\times {{10}^{-12}}}$

$\Rightarrow E=1.92\times {{10}^{-10}}N/C$

Thus, it can be concluded that the electric field between the plates is $1.92\times {{10}^{-10}}N/C$.

11. An Oil Drop of $12$ Excess Electrons Is Held Stationary Under a Constant Electric Field of $2.55\times {{10}^{4}}N{{C}^{-1}}$ in Millikan's Oil Drop Experiment. the Density of the Oil Is $1.26gm/c{{m}^{3}}$. Estimate the Radius of the Drop. $\left( g=9.81m{{s}^{-2}},e=1.60\times {{10}^{-19}}C \right)$.

The number of excess electrons on the oil drop, \[n=12\] 

Electric field intensity, $E=2.55\times {{10}^{4}}N{{C}^{-1}}$ 

The density of oil, $\rho =1.26gm/c{{m}^{3}}=1.26\times {{10}^{3}}kg/{{m}^{3}}$ 

Acceleration due to gravity, $g=9.81m{{s}^{-2}}$ 

Charge on an electron $e=1.60\times {{10}^{-19}}C$

Radius of the oil drop $=r$ 

Here, the force (F) due to electric field E is equal to the weight of the oil drop (W).

$Ene=\frac{4}{3}\pi {{r}^{2}}\rho \times g$ 

$q$ is the net charge on the oil drop $=ne$

$m$ is the mass of the oil drop $=\text{Volume of the oil drop}\times \text{Density of oil}$\[=\frac{4}{3}\pi {{r}^{3}}\times p\]

Therefore, radius of the oil drop can be calculated as 

\[r=\sqrt{\frac{3Ene}{4\pi \rho g}}\]

\[\Rightarrow r=\sqrt{\frac{3\times 2.55\times {{10}^{4}}\times 12\times 1.6\times {{10}^{-19}}}{4\times 3.14\times 1.26\times {{10}^{3}}\times 9.81}}\]

\[\Rightarrow r=\sqrt{946.09\times {{10}^{-21}}}\]

\[\Rightarrow r=9.72\times {{10}^{-10}}m\]

Therefore, the radius of the oil drop is $9.72\times {{10}^{-10}}m$.

12. In a Certain Region of Space, Electric Field Is Along the Z-Direction Throughout. the Magnitude of Electric Field Is, However, Not Constant but Increases Uniformly Along the Positive Z-Direction, at the Rate of ${{10}^{5}}N{{C}^{-1}}$ Per Meter. What Are the Force and Torque Experienced by a System Having a Total Dipole Moment Equal to ${{10}^{-7}}Cm$ in the Negative Z-Direction?

It is known that the dipole moment of the system, $P=q\times dl=-{{10}^{-7}}Cm$

Also, the rate of increase of electric field per unit length is given as

 $\frac{dE}{dl}={{10}^{5}}N{{C}^{-1}}$

Now, the force (F) experienced by the system is given by $F=qE$

$F=q\frac{dE}{dl}\times dl$

$F=P\frac{dE}{dl}$

$\Rightarrow F=-{{10}^{-7}}\times {{10}^{5}}$

$\Rightarrow F=-{{10}^{-2}}N$ 

Clearly, the force is equal to $-{{10}^{-2}}N$ in the negative z-direction i.e., it is opposite to the direction of electric field. 

Thus, the angle between electric field and dipole moment is equal to \[180{}^\circ \].

Now, the torque is given by $\tau =PE\sin \theta $

𝜏 = PEsin180 0

Therefore, it can be concluded that the torque experienced by the system is zero.

a) A Conductor With a Cavity as Shown in the Fig. 1.36(a) Is Given a Charge $Q$. Show That the Entire Charge Must Appear on the Outer Surface of the Conductor. 

Firstly, let us consider a Gaussian surface that is lying within a conductor as a whole and enclosing the cavity. Clearly, the electric field intensity E inside the charged conductor is zero.

Now, let \[q\] be the charge inside the conductor and ${{\varepsilon }_{0}}$, the permittivity of free space.

According to Gauss's law,

Flux is given by

 $\phi =\overrightarrow{E}.ds=\frac{q}{{{\varepsilon }_{0}}}$ 

Here, $\phi =0$  as $E=0$ inside the conductor

Clearly, 

$0=\frac{q}{8.854\times {{10}^{-12}}}$

$\Rightarrow q=0$ 

Therefore, the charge inside the conductor is zero.

And hence, the entire charge $Q$ appears on the outer surface of the conductor.

b) Another Conductor B With Charge $q$ is Inserted Into Cavity Keeping B Insulated from A. Show That the Total Charge on the Outside Surface of A is $Q+q$ [Fig. 1.36 (b)].

The outer surface of conductor A has a charge of $Q$.

It is given that another conductor B, having a charge \[+q\] is kept inside conductor A and is insulated from the conductor A.

Clearly, a charge of \[-q\] will get induced in the inner surface of conductor A and a charge of \[+q\] will get induced on the outer surface of conductor A.

Therefore, the total charge on the outer surface of conductor A amounts to $Q+q$.

C) a Sensitive Instrument is to Be Shielded from the Strong Electrostatic Fields in Its Environment. Suggest a Possible Way.

A sensitive instrument can be shielded from a strong electrostatic field in its environment by enclosing it fully inside a metallic envelope. 

Such a closed metallic body provides hindrance to electrostatic fields and thus can be used as a shield.

14. A Hollow Charged Conductor Has a Tiny Hole Cut Into Its Surface. Show That the Electric Field in the Hole Is $\left[ \frac{\sigma }{2{{\varepsilon }_{0}}} \right]\overset{\wedge }{\mathop{n}}\,$, where $\overset{\wedge }{\mathop{n}}\,$ is the Unit Vector in the Outward Normal Direction, and $\sigma $ Is the Surface Charge Density Near the Hole.

Firstly, let us consider a conductor with a cavity or a hole. It is known that the electric field inside the cavity is zero. 

Let us assume E to be the electric field just outside the conductor, $q$ be the electric charge, $\sigma $ be the charge density, and ${{\varepsilon }_{0}}$, the permittivity of free space.

We know that charge $\left| q \right|=\sigma \times d$ 

Now, according to Gauss's law,

$\phi =E.ds=\frac{\left| q \right|}{{{\varepsilon }_{0}}}$

$E.ds=\frac{\sigma \times d}{{{\varepsilon }_{0}}}$

$\therefore E=\frac{\sigma }{{{\varepsilon }_{0}}}\overset{\wedge }{\mathop{n}}\,$

where $\overset{\wedge }{\mathop{n}}\,$ is the unit vector in the outward normal direction.  

Thus, the electric field just outside the conductor is $\frac{\sigma }{{{\varepsilon }_{0}}}\overset{\wedge }{\mathop{n}}\,$. Now, this field is actually a superposition of the field due to the cavity ${{E}_{1}}$ and the field due to the rest of the charged conductor ${{E}_{2}}$. These electric fields are equal and opposite inside the conductor whereas equal in magnitude as well as direction outside the conductor. Clearly,

${{E}_{1}}+{{E}_{2}}=E$

${{E}_{1}}={{E}_{2}}=\frac{E}{2}=\frac{\sigma }{2{{\varepsilon }_{0}}}\overset{\wedge }{\mathop{n}}\,$

Therefore, the electric field in the hole is  $\frac{\sigma }{2{{\varepsilon }_{0}}}\overset{\wedge }{\mathop{n}}\,$. 

Hence, proved.

15. Obtain the Formula for the Electric Field Due to a Long Thin Wire of Uniform Linear Charge Density $\lambda $ Without Using Gauss's Law. (hint: Use Coulomb's Law Directly and Evaluate the Necessary Integral)

Firstly, let us take a long thin wire XY as shown in the figure below. This wire is of uniform linear charge density $\lambda $ .

Now, consider a point A at a perpendicular distance l from the mid-point O of the wire as shown in the figure below:

Consider E to be the electric field at point A due to the wire.

Also consider a small length element $dx$ on the wire section with $OZ=x$ as shown.

Let $q$ be the charge on this element.

Clearly, $q=\lambda dx$ 

Now, the electric field due to this small element can be given as

$dE=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\lambda dx}{{{\left( AZ \right)}^{2}}}$ 

However, $AZ=\sqrt{{{1}^{2}}+{{x}^{2}}}$ 

\[\therefore dE=\frac{\lambda dx}{4\pi {{\varepsilon }_{0}}\left( {{1}^{2}}+{{x}^{2}} \right)}\]

Now, let us resolve the electric field into two rectangular components. Doing so, $dE\cos \theta $ is the perpendicular component and $dE\sin \theta $ is the parallel component.

When the whole wire is considered, the component $dE\sin \theta $ gets cancelled and only the perpendicular component $dE\cos \theta $  affects the point A.

Thus, the effective electric field at point A due to the element $dx$ can be written as 

$d{{E}_{1}}=\frac{\lambda dx\cos \theta }{4\pi {{\varepsilon }_{0}}\left( {{l}^{2}}+{{x}^{2}} \right)}$    ....(1) 

Now, in $\Delta AZO$, we have

$\tan \theta =\frac{x}{l}$

$x=l\tan \theta \text{          }......\text{(2)}$ 

On differentiating equation (2), we obtain

$dx=l{{\sec }^{2}}d\theta \text{    }......\text{(3)}$ 

From equation (2)

${{x}^{2}}+{{l}^{2}}={{l}^{2}}+{{l}^{2}}{{\tan }^{2}}\theta $

$\Rightarrow {{l}^{2}}\left( 1+{{\tan }^{2}}\theta  \right)={{l}^{2}}{{\sec }^{2}}\theta $

$\Rightarrow {{x}^{2}}+{{l}^{2}}={{l}^{2}}{{\sec }^{2}}\theta \text{         }.....\text{(4)}$

Putting equations (3) and (4) in equation (1), we obtain

$d{{E}_{1}}=\frac{\lambda l{{\sec }^{2}}d\theta }{4\pi {{\varepsilon }_{0}}\left( {{l}^{2}}{{\sec }^{2}}\theta  \right)}\cos \theta $

$\therefore d{{E}_{1}}=\frac{\lambda \cos \theta d\theta }{4\pi {{\varepsilon }_{0}}l}\text{       }.....\text{(5)}$

Now, the wire is taken so long that ends from $-\frac{\pi }{2}$ to $+\frac{\pi }{2}$.

Therefore, by integrating equation (5), we obtain the value of field ${{E}_{1}}$ as

\[\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{d{{E}_{1}}}=\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\frac{\lambda }{4\pi {{\varepsilon }_{0}}l}\cos \theta d\theta }\]

\[\Rightarrow {{E}_{1}}=\frac{\lambda }{4\pi {{\varepsilon }_{0}}l}\times \text{2}\]

\[\Rightarrow {{E}_{1}}=\frac{\lambda }{2\pi {{\varepsilon }_{0}}l}\]

Thus, the electric field due to the long wire is derived to be equal to \[\frac{\lambda }{2\pi {{\varepsilon }_{0}}l}\].

16. it Is Now Believed That Protons and Neutrons (which Constitute Nuclei of Ordinary Matter) Are Themselves Built Out of More Elementary Units Called Quarks. a Proton and a Neutron Consist of Three Quarks Each. Two Types of Quarks, the So Called 'up Quark (denoted by $u$)  of Charge $\left( +\frac{1}{2} \right)e$ and the 'down' Quark (denoted by $d$) of Charge $-\left( \frac{1}{3} \right)e$ Together With Electrons Build up Ordinary Matter. (Quarks of Other Types Have Also Been Found Which Give Rise to Different Unusual Varieties of Matter.) Suggest a Possible Quark Composition of a Proton and Neutron.                       

It is known that a proton has three quarks. Let us consider \[n\] up quarks in a proton, each having a charge of $+\left( \frac{2}{3}e \right)$. 

Now, the charge due to \[n\] up quarks $=\left( \frac{2}{3}e \right)n$ 

The number of down quarks in a proton $=3-n$ 

Also, each down quark has a charge of $-\frac{1}{3}e$ 

Therefore, the charge due to $\left( 3-n \right)$ down quarks $=\left( -\frac{1}{3}e \right)\left( 3-n \right)$

We know that the total charge on a proton $=+e$ 

$e=\left( \frac{2}{3}e \right)n+\left( -\frac{1}{3}e \right)\left( 3-n \right)$

$\Rightarrow e=\left( \frac{2ne}{3} \right)-e+\frac{ne}{3}$

$\Rightarrow 2e=ne$

$\Rightarrow n=2$

Clearly, the number of up quarks in a proton, $n=2$

Thus, the number of down quarks in a proton $=3-n=3-2=1$ 

Therefore, a proton can be represented as $uud$.

A neutron is also said to have three quarks. Let us consider \[n\] up quarks in a neutron, each having a charge of $+\left( \frac{2}{3}e \right)$ .

It is given that the charge on a neutron due to \[n\] up quarks $=\left( +\frac{3}{2}e \right)n$ 

Also, the number of down quarks is $\left( 3-n \right)$, each having a charge of $=\left( -\frac{3}{2} \right)e$

Thus, the charge on a neutron due to $\left( 3-n \right)$ down quarks $=\left( -\frac{1}{3}e \right)\left( 3-n \right)$

Now, we know that the total charge on a neutron $=0$ 

$0=\left( \frac{2}{3}e \right)n+\left( -\frac{1}{3}e \right)\left( 3-n \right)$

$\Rightarrow 0=\left( \frac{2ne}{3} \right)-e+\frac{ne}{3}$

$\Rightarrow e=ne$

$\Rightarrow n=1$

Clearly, the number of up quarks in a neutron, $n=1$

Thus, the number of down quarks in a neutron $=3-n=2$ 

Therefore, a neutron can be represented as $udd$.

a) Consider an Arbitrary Electrostatic Field Configuration. A Small Test Charge Is Placed at a Null Point (i.e., where \[\mathbf{E}=\mathbf{0}\]) Of the Configuration. Show That the Equilibrium of the Test Charge Is Necessarily Unstable.

Firstly, let us assume that the small test charge placed at the null point of the given setup is in stable equilibrium. 

By stable equilibrium, it means that even a slight displacement of the test charge in any direction will cause the charge to return to the null point as there will be strong restoring forces acting around it. 

This further suggests that all the electric lines of force around the null point act inwards and towards the given null point. 

But by Gauss law, we know that the net electric flux through a chargeless enclosing surface is equal to zero. This truth contradicts the assumption which we had started with. Therefore, it can be concluded that the equilibrium of the test charge is necessarily unstable.

b) Verify this Result for the Simple Configuration of Two Charges of the Same Magnitude and Sign Placed at a Certain Distance Apart.

When we consider this configuration setup with two charges of the same magnitude and sign placed at a certain distance apart, the null point happens to be at the mid-point of the line joining these two charges. 

As per the previous assumption, the test charge, when placed at this mid-point will experience strong restoring forces when it tries to displace itself. 

But when the test charge tries to displace in a direction normal to the line joining the two charges, the test charge gets pulled off as there is no restoring force along the normal to the line considered. 

Since stable equilibrium prioritizes restoring force in all directions, the assumption in this case also gets contradicted. 

18. A Particle of Mass $m$ and Charge $\left( -q \right)$ Enters the Region Between the Two Charged Plates Initially Moving Along X- Axis With Speed $vx$ (like particle 1 in Fig 1.33). The Length of Plate is $L$ and a Uniform Electric Field $E$ Is Maintained Between the Plates. Show That the Vertical Deflection of the Particle at the Far Edge of the Plate is $\frac{qE{{L}^{2}}}{2m{{v}_{x}}^{2}}$.

Compare This Motion with Motion of a Projectile in Gravitational Field Discussed in Section 4.10 of Class Xi Textbook of Physics.

The charge on a particle of mass $m=-q$

Velocity of the particle $=vx$ 

Length of the plates $=L$ 

Magnitude of the uniform electric field between the plates $=E$ 

Mechanical force, $F=\text{ Mass }(m)\times \text{Acceleration }(a)$ 

Thus, acceleration, $a=\frac{F}{m}$

However, electric force, $F=qE$ 

Therefore, acceleration, $=\frac{qE}{m}$  .........(1)

Here, the time taken by the particle to cross the field of length $L$ is given by,

$t=\frac{\text{Length of the plate}}{\text{Velocity of the plate}}=\frac{L}{{{v}_{x}}}$  ......(2)

In the vertical direction, we know that the initial velocity, $u=0$ 

Now, according to the third equation of motion, vertical deflection $s$ of the particle can be derived as

$s=ut+\frac{1}{2}a{{t}^{2}}$

$\Rightarrow s=0+\frac{1}{2}\left( \frac{qE}{m} \right){{\left( \frac{L}{{{v}_{x}}} \right)}^{2}}$ 

$\Rightarrow s=\frac{qE{{L}^{2}}}{2m{{v}_{x}}^{2}}$   .....(3)

Therefore, the vertical deflection of the particle at the far edge of the plate is $\frac{qE{{L}^{2}}}{2m{{v}_{x}}^{2}}$ 

On comparison, we can see that this is similar to the motion of horizontal projectiles under gravity.

Refer to the below-given table to know the important topics and subtopics of CBSE Class 12 Physics Chapter 1 - Electric Charges and Fields, and plan your preparation strategy accordingly with Electric Charges and Fields Class 12 Important Questions.

Important Question of Physics Class 12 Chapter 1 - Free PDF Download

Students who are weak in studies might consider science difficult for them. They might be incapable of understanding the different concepts that are involved in science. Chapter 1 of science in class 12 consists of many advanced studies that a student can understand only if he/she has a strong core knowledge. Considering the situation, the best way to face this problem is practice. When a student practices different questions regularly, he/she gains quite a fair amount of knowledge which will help them in scoring good marks.

If students are confused and can't find the questions they are required to practice, then they must refer to electric charges and fields class 12 important questions. Chapter 1 physics class 12 important questions are available in a pdf format on the Vedantu website. Students can download it for free and use it in different stages of their life. Students must not ignore this because this can help them to score a lot of marks.

Electric Charges and Fields Important Questions for Class 12 Physics

While practising the important question of chapter 1 physics class 12, students learn a lot of things that strengthen their core knowledge, and they become capable of scoring good marks in the exams. Class 12 physics ch 1 important questions is a saviour for those students who have a weak core and do not understand the concepts of the chapter. Some of the knowledge that students gain during practising physics Class 12 Physics Ch 1 Important Questions are as a follows:

Electric Charge

The term 'electricity' got derived from a Greek word Elektron, which means amber. The magnetic and electric forces present inside matter, atoms, and molecules are used to determine their properties. There remain only two kinds of the entity which are known as electric charge.

An experiment was conducted to analyze the electric charge. This experiment suggested that there are two kinds of electrification which are: (i) like charges repel from each other and (ii) unlike charges seems to attract each other. There's a property which differentiates these two kinds of charges which are known as the polarity of charge.

Electric charge is considered as the physical property of matter which causes it to experience a force when it gets placed on an electromagnetic field. Electric charges are classified into two types which is negative and positive. The negatively charged matter is known as an electron, whereas a positively charged matter is known as a proton.

Properties of Electric Charge

Some fundamental properties of the electric charge are as follows:

Quantization - This property of electric charge suggests that the total charge of a body can represent the integral multiple of a basic quantum of charge.

Additive - This property of electric charge suggests that the total charge of a body can represent the total sum of all the singular charges that seem to be acting on the system.

Conservation - This property of electric charge suggests that the total or the whole charge of a particular system remains unaffected with time. In simple words, when an object gets charged due to some friction, a transfer of charge from one object to another object occurs. A charge is a thing which can't be destroyed or created.

Electric Field

An electric field is considered as the electric force that is present in one unit of charge. The direction of the field is considered to be the same as the direction of the force, and it is expected to exert on a positive test charge. When the electric field is radially outward, then it is considered as a positive charge, and when the electric field is radially inward, then it is considered as a negative charge. The SI unit of the electric field is volt per meter.

In simple words, we can define the electric field as a physical field which can be found surrounding each electric charge and is responsible for exerting force on all the other charges present on the field. Electric fields are believed to be originated from electric or from time-varying magnetic fields. Both electric fields and magnetic fields are considered as the manifestations of the electromagnetic force, which is one of the four fundamental forces that exist in nature or the environment.

Properties of Electric Field

The following are considered as the properties of an electric field:

The lines that are present in the electric field never intersect with each other.

The lines in the electric field are perpendicular to the surface charge.

When the lines are closely held together, the field is strong, but when the lines are loose and are not close together, the field is weak.

The number of lines in the field is directly proportional to the magnitude of the charge.

The lines in the electric field seem to start from a positive charge and end with a negative charge.

When the charge is single, then they seem to start and end at infinity.

The line curves of the electric field are continuous in a charge-free region.

All this above information are basic theories of the chapter 'Electric charges and fields'. More advanced theories and concepts are there in the chapter that the students will encounter while going through the chapter. Students must practice the class 12th physics chapter 1 important questions regularly to understand the advance theories and easily so that he/she will be capable of scoring good marks in the exams. If students ignore to practice Electric Charges and Fields Class 12 Important Questions , then they will face a lot of problems. Especially weak students who don't have a strong core knowledge will be the highest sufferers.

Physics Class 12 Chapter 1 Important Questions

Some of the electric charges and fields class 12 important questions that are most likely to come in the exams are as follows:

What do you understand by term Electric charge? Explain it with the help of an experiment.

What do you understand by the term conductors?

What do you understand by the terms insulators?

What is the difference between conductors and insulators?

State the properties of an electric charge.

Explain Coulomb's Law. State some examples and experiments.

Discuss the forces that are present between the multiple charges.

State the properties of the electric field lines.

What is the meaning of the term electric flux?

What do you understand by the term electric dipole?

State the SI unit of the electric field. Explain how it got derived.

Electric field lines cannot intersect with each other. Explain why.

Why are the electric field lines perpendicular to the surface centre?

Physics Class 12 Chapter 1 Important Questions Related Links

Discover the essential topics of Physics Class 12 Chapter 1, covering everything from Electric Charges to Electric Fields. Whether you're studying for exams or simply exploring physics, this list offers a clear roadmap to understanding fundamental principles.

Electric Charges and Fields

Electric Field Intensity

Electric Field Formula

Electric Field Lines Characteristics 

Basic Properties Electrical Charge

Electric Charge and Static Electricity

Benefits of Electric Charges and Fields Important Questions with Answers PDF

Students who are weak in science get a lot of advantage and knowledge by regularly practising the Physics Class 12 Chapter 1 Important Questions PDF . Its acts like a guide for every student. So some of the benefits of referring to important questions for class 12 physics chapter 1 are as follows:

The questions which are listed are selected considering the format and the syllabus imposed by the CBSE board because any deviations from these can cost students a lot of marks.

Electric Charges and Fields Important Questions with Answers PDF are prepared by some expert individuals who are professionals in the field and have worked for years.

The selection of question is made keeping in mind the intellectual capability of students so that they can attain every question.

The questions that are included in the list are most likely to come in the final exams, thus preparing the students better.

For Further Assistance Watch our Master Teacher Anupam Sir Explaining Electric Charges and Field | Class 12 Physics  

Also watch Electric Charges & Fields Full Chapter in 60 Minutes Class 12 Physics Ch 1 One Shot by Anupam Sir

Students who are unable to understand the concepts of science must consider referring to Physics Class 12 Chapter 1 Important Questions PDF so that they can score the highest possible marks in the exams. They will be clear of all the doubts.

Important Related Links for CBSE Class 12 Physics 

FAQs on Electric Charges and Fields Class 12 Important Questions with Solutions

1. Where can I find Electric Charges and Fields Class 12 Important Questions?

Vedantu provides selected questions for each chapter of the NCERT textbook which are important for exam preparation. Students can avail the important questions PDF file for Class 12 Physics Chapter 1 Electric Charges and Fields on Vedantu’s platform. The free PDF version allows students to download these files anytime, anywhere. The list of important questions is picked with great efforts by expert Physics tutors to help students in scoring well in the paper. They also offer the solutions as per the board guidelines and exam pattern. Students must practice these questions in order to revise all types of questions and get an idea of what might be asked in exams.

2. Why should I practice Vedantu’s important questions for Class 12 Physics Chapter 1?

Many students find it challenging to score well in Class 12 Physics paper. However, with a thorough understanding of concepts and regular practice, scoring in the Physics paper won’t be difficult. For the better practice of Class 12 Physics Chapter 1 as well as other chapters, students can download the free PDF file of important questions from Vedantu’s site. These questions are added as per the exam pattern and the latest syllabus. This will ensure that you have better coverage on the concepts as well as a fair idea of what types of questions to expect in the exam. You can also find solutions for these questions provided by experts in case they are having trouble solving them.

3. What are some of the important questions for Class 12 Physics Chapter 1?

Following are some of the important questions for Class 12 Physics Chapter 1 - Electric Charges and Fields:

A charged rod P attracts rod R whereas P repels another charged rod Q. What type of force is developed between Q and R?

Define one coulomb.

No two electric lines of force can intersect each other? Why?

Explain why two field lines never cross each other at any point.

What is the force between two small charged spheres having charges of 2 x 10 -7 C and 3 x 10 -3 C placed 30 cm apart in the air?

For solutions to these questions as well as more important questions on Class 12 Physics Chapter 1, students can refer to Vedantu’s important questions for Class 12 Physics Chapter 1.

4. What are some of the important contents of Chapter 1 Electric Charges and Fields of Class 12 Physics?

Class 12 Physics Chapter 1 Electric Charges and Fields consists of questions on the linear charge density of an infinite line charge and many other questions regarding the electric field. In this chapter, students will be solving questions on how to calculate forces between two charged particles which are kept at a certain distance from each other. Students will be learning about the theory behind the charge appearing in materials due to rubbing against each other just like when we rub a glass rod with a silk cloth or a dry comb on our hair. There are many other interesting concepts in this chapter.

5. How are Electric Charges and Fields Class 12 Important Questions beneficial for students?

Important questions of Class 12 Physics Chapter 1 available at Vedantu app and website can be extremely useful for the students of Class 12. They can use the important questions for their board exam preparation and can also prepare for their entrance exam. The important questions can help them to understand the concepts easily because all important questions are prepared using basic and simple language that makes it easy for the students to understand the facts. 

6. How can I download important questions for Chapter 1- Electric Charges and Fields of Class 12 Physics from Vedantu?

Students can download the Chapter 1- Electric Charges and Field of Class 12 Physics important questions from Vedantu free of cost. They have to visit the website and search for important questions of Chapter 1 of Class 12 Physics then click on the download link. They can save the PDF file on their computers and can use the important questions to prepare for their final exams and entrance exam. All questions are important for the board exams and NEET and JEE Exams.

7. What are the main concepts students will study in Chapter 1 of Class 12 Physics?

Students will study different concepts related to Electric Charges and Fields in Class 12 Physics Chapter 1. They will study Electrostatic capacitance and potential, current electricity, the relation between moving charges and magnetism, electromagnetic induction, alternating current, and electromagnetic waves. Students must understand all the concepts given and can refer to the important questions given on Vedantu for a clear understanding of the concepts.

8. What is an electric charge according to Chapter 1 of Class 12 Physics?

Electric charge is a physical property of matter. The matter experiences a force when it is placed in an electromagnetic field. Electric charges can be positive or negative. The particle that carries a negative charge is called an electron and the particle that carries a positive charge is called a proton. Many experiments were done for analyzing electric charge. The experiment revealed that like charges move away from each other and unlike charges come closer to each other. 

9. Write a short note on the electric field related to Chapter 1 of Class 12 Physics?

The electric field is a field that surrounds an electric charge. It exerts a force on other charges present in the field. The direction of the force exerted on the charge is the same as the direction of the field. An electric field is believed to be a result of electromagnetic force. If the direction of the force is outward it is considered positive and if the direction of the force is inward it is considered negative. 

10. What are the important topics in electric charges and fields 12?

Important topics in Electric Charges and Fields Class 12 include electric charge, Coulomb's law, electric field and electric field lines, Gauss's law, electric potential, capacitance, and conductors and insulators.

11. What is charge class 12 questions?

Questions on charge in Class 12 typically cover topics such as the properties of charge, conservation of charge, Coulomb's law, and the behavior of charged particles in electric fields.

12. Is the electric field due to a charge configuration with total charge zero necessarily?

No, the electric field due to a charge configuration with total charge zero is not necessarily zero. The electric field depends on the distribution of charges, and even if the total charge is zero, the electric field may not be zero due to the arrangement of charges.

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Physics Class 12 Chapter 1 Important Questions

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With Class 12 Physics Chapter 1 Important Questions, students will get detailed and authentic solutions to questions regarding electric charges and fields. These Class 12 Physics Chapter 1 Important Questions will prepare students for their exams. After studying these important questions , students will be able to solve many CBSE sample papers .

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In addition, these important question notes also contain important formulas and CBSE extra questions that can help students test their understanding.

CBSE Class 12 Physics Chapter 1 Important Questions

Study important questions for class 12 physics chapter 1 – electric charges and fields.

Electric charge refers to a matter’s basic physical property that results in it experiencing a force when it is kept in a magnetic or an electric field. The interaction between the charges causes an electromagnetic force. Here are some important questions that students can that study to prepare them better for their examinations.

Short Questions and Answers (1 Mark)

1.  name the physical quantity that has these s.i. units (i) cm (ii) n c.

Ans. (i) Cm is the S.I. unit of electric dipole moment.

(i) NC is the S.I. unit of electric field intensity.

2. Which physical quantity has the S.I. unit of JC -1 ? Also, state whether it is a vector or a scalar quantity.

Ans. JC-1 is the S.I. unit of electric potential, and it is a scalar quantity.

3. What do you mean by one Coulomb?

Ans. 1 Coulomb refers to the charge on a body if it receives a force of either attraction or repulsion of 9109N  from another equal charge when they have a distance of 1m between them.

4. Will the force existing between two point charges change if the medium’s dielectric constant in which they are present increases?

Ans. The dielectric constant of a medium is presented by the following equation

k=FVFM=Force between the charges in vacuumForce between two charges in medium

According to the expression given above, we can clearly observe that k increase and FM decreases.

5. A charge ‘q’ is positioned at the middle of a cube of side l. What is the electric flux travelling through two opposing faces of the cube? 

Ans.  E= q30

Short Questions and Answers (2 Marks)

Q.1. Why can’t two electric lines of force intersect each other?

Ans. This can be understood through an example. Let’s imagine if they intersect, then at the place of contact, you can draw two tangents from it.

These two tangents are intended to depict two directions of electric field lines, which, at a given spot, is not achievable.

Q.2. Why is the quantization of electric charge when working with macroscopic charge irrelevant?

Ans. When dealing with large-scale and macroscopic charges, the charges used are immense in quantity as compared to the size of the electric charge. That’s why the quantization of electric charge is irrelevant on a macroscopic scale. Therefore, it is ignored, and it is assumed that the electric charge is continuous.

Q.3. Calculate the dipole moment of the dipole when an electric dipole is held at 30° to the uniform electric field of 10 4 N/C and experiences a torque of 9 10 -26 Nm.

Ans. The details are given as follows.

t=910-26 Nm

P denotes the dipole moment that is required to be calculated.

The torque is presented by =PEsin

P= 910-26104sin 30° = 910-2610-421

P= 1810-30 Cm

Q.4. When a glass rod is brushed with a silk cloth, charges develop on both objects. A similar phenomenon is seen with numerous different pairs of entities. Explain how this occurs with the law of conservation of charge.

Ans. When two objects are rubbed against each other, there is a generation of charges equal in magnitude, but contrary in nature to the bodies that are rubbed. It is also noted that during such an occurrence, charges are produced in pairs. This phenomenon of charging is referred to as charging by friction.

The net charge on the system of both bodies equals zero. This is because the same number of opposite charges in both bodies destroy each other. When a silk cloth is rubbed on a glass rod, opposite-natured charges appear on both those bodies. Thus, this event agrees with the law of conservation of energy. A similar phenomenon can also be observed with numerous different pairs of bodies too.

Q.5. What does the phrase ‘electric charge of a body is quantized’ mean?

Ans: The phrase ‘electric charge of a body is quantized’ says that only an integral (1, 2, 3, 4, …, n) number of electrons may be passed from one body to another. This further implies that charges are not transmitted in fractions. Hence, a body possesses its entire charge only in integral multiples of electric charges.

Q.6. What will the net charge within the box be when the net outwards flux on the box’s surface is 8.0 10³ Nm 2 /C ?

Ans. The information given is as follows

Net outward flux surface of the box 8.0 10³Nm2/C

If a body contains the net charge q, we can indicate the flux by =q0

0=Permittivity of free space=8.854 10-12 N-1 C2 m-2

Hence, the charge q can denote by q=0

⇒q= 8.854 10–12 8.0 10³

⇒q= 7.08 10-8

Hence, 0.07C will be the net charge inside the box.

Q.7. An electric field line is a continuous curve, meaning that a field line cannot have unexpected breaks. Why not?

Ans. An electrostatic field line is a constant curve since it is understood that a charge experiences a continuous force when tracked in an electrostatic field. In addition, the field line cannot have unexpected breaks since the charge moves continuously and cannot jump from one place to another.

Q.8. What is an electric dipole moment, and what is its S.I Unit?

Ans. The product of the magnitude of either dipole length or change is called the electric dipole moment of an electric dipole.

S.I. unit of dipole ( p ) is coulomb metre (Cm).

Q.9. A particle having mass m and charge q is discharged from its rest in a uniform electric field of intensity denoted by letter E. Find out the kinetic energy achieved by the particle after travelling a distance.

Ans. In an electric field, E and the electrostatic force is applied to a change can be written as

F = qE          …………………….(1)

We also know Newton’s second law of motion is

F = ma          …………………….(2)

From (1) and (2)

a = qEm           …………………….(3)

The third equation of motion is

v² – u² = 2as

Since the charged particle is at rest

v² = 2as       ……………………. (4)

The kinetic energy is given by

KE = 12mv²        ………………….(5)

Substituting (4) in (5), we get,

KE = 12m(2as)=mas   ……………(6)

Substituting (3) in (6) to get,

Therefore, the kinetic energy that is achieved by the particle with a charge of q that is moving a distance s in the electric field E is denoted by

Q.10. What will the force between two small spheres that have 2 10 -7 C and 3 10 -7 C be, if they are suspended in the air and have 30 cm of distance between them?

Ans. The charge of the 1st and 2nd spheres is 210-7 C and 310-7C

The distance r is = 30 cm = 0.3m

The electrostatic force that exists between the spheres can be denoted as

F= 140.q1q2r2

0= permittivity of free space and 140= 9109 Nm² C-2

Force, F=9109210-7310-70.32= 610-3 N

A repulsive force will exist between the charges as they are of similar nature.

Q.11. A sphere S 1 having radius R 1 surrounds a charge Q. Another concentric sphere S 2 of radius R 2 (R 2 ﹥R 1 ) with no additional charge between S 1 and S 2 , find the ratio of electric flux between S 1 and S 2 .

Ans. We can recall that according to Gauss’s law, the expression for electric flux going through a surface q is

Here, 0 denotes the permittivity of the medium.

The electric flux that is going through sphere S1 is denoted by

s1= Q0       …………….(1)

We know that there is no additional charge exists between the two spheres. The flux existing between S2 is given by

s2= Q0       ……………..(2)

Hence the required ratio will be 1:1

Q.12. If a polythene piece is rubbed with wool and results in a negative charge of 3 10 -7 , answer the following.

• Calculate how many electrons will be moved, noting which item they will come from and go to.
• Will there be a transfer of mass from wool to polythene?

Ans. (a) When a piece of polythene is rubbed against the wool, a certain number of electrons are passed on from wool to polythene. So, this results in wool becoming positively charged, while losing electrons and polythene becomes negatively charged when it gains the electrons.

Charge on the polyethene piece,

And the charge of electrons,

e =-1.610-19 C

Now let’s imagine n to be the number of electrons that are transferred from wool to polythene, after that, using the property of quantization of charge, we can say

n = -310-7-1.610-19

∴n = 1.871012

Hence, 1.871012 will be the number of electrons that are transferred from wool to polythene.

(b) Yes, there will be a transfer of electrons from wool to polythene, and mass will also be transferred.

For instance, m is the mass being passed on in the given case, and me is the mass of the electron then,

m = 9.110-311.851012

∴m = 1.70610-18

Therefore, we can observe that only a small amount of mass is passed on from wool to polythene.

Q.13. Imagine there is a uniform electric field E = 3 10 3 îN/C. Find out the following.

• The flux of this electric field is through a square of side 10 cm, the plane of which is parallel to the y-z plane.
• The flux through the same square if the 60 ° is formed by the normal plane with the x-axis.

Ans. (a) The information given is as follows.

Electric field intensity, E = 33103 îN/C

The magnitude of the electric field intensity, E = 3103 N/C

Side of the square, a=10cm = 0.1m

Area of the square, A = a² = 0.01m²

Since the square’s plane is parallel to the y-z plane, the normal plane would move in the x direction. So, the angle between the normal plane and the electric field would be, = 0°

When we substitute the values, we get,

= 310³0.01cos 0°

∴ = 3-Nm²/c

The net flux through the above-mentioned surface will be = 3-Nm²/c

(b) When the angle is 60° with the x-axis, the flox through the given surface will be,

= 310³0.01cos 60°

∴ = 15Nm²/C

So the flux here would be 15Nm²/C.

Q.14. The distance between a point charge +10 C is 5 cm above the square’s centre, having a side of 10cm, as given in the figure below. What will the magnitude of electric flux through the square be? (Hint: Imagine one face of a cube to be a square with an edge of 10cm.)

Ans. Imagine one face of a cube to be a square having 10cm with a charge q in its centre. Now, as explained by Gauss’s theorem for a cube, total electric flux is through its every six faces.

total =  q0

The electric flux passing through one cube’s face can be given by, = total6

The permittivity of free space, 0 = 8.85410-12N-1C²m-2

The net charge enclosed, q = 10C = 1010-6C

If we substitute the values in the question

= 161010-6C8.854105Nm2C-1

∴ = 1.88105Nm2C-1

Hence, the electric flux through the square will be 1.88105Nm2C-1.

Q.15. A spherical Gaussian surface with a 10 cm radius that is centred on a point charge experiences an electric flux of  -0.1 10 3 Nm 2 /C. Answer the following questions.

• How much flux is capable of passing through the surface if the radius of the Gaussian surface is doubled?
• What will the value of the point charge be?

Ans (a). Electric flux caused by the point charge is -0.1103Nm2/C.

The Gaussian surface’s radius that encloses the point charge is

r = 10.0 cm.

The net charge contained by the surface, as determined by Gauss’ law, determines how much electric flux pierces through the surface.

It is unaffected by the size of the hypothetical surface that is expected to encapsulate this charge.

The flux travelling through the surface, or 103Nm² / C, remains the same if the radius of the Gaussian surface is doubled.

(b) The electric flux is denoted by the relation,

q = net charge encircled by the spherical surface

Permittivity of free space, 0= 8.85410-12C²m-2

If we substitute the value,

q = -1.010³8.85410-12 =-8.85410-9C

∴ = -8.854nC

Hence, -8.854nC will be the value of the point charge.

Q.16. The surface charge density of a uniformly charged conduction sphere having a diameter of 2.4m is 80.0 C/m². Answer the following questions.

• Find the charge that the sphere would have.
• What will the total electron flux leaving the surface of the sphere be?

Ans. (a) The diameter of the sphere is

The radius of the sphere is

The surface charge density,

= 80.0C/m² = 8010-6 C/m²

The total charge that will be present on the surface of the sphere will be

Q = Charge densitySurface area

Q = 4r²+ 8010-6 43.14(1.2)²

∴Q = 1.44710-3C

Hence, the charge present on the surface will be 1.44710-3C.

(b) As given by the Gauss’s law, the total electric flux (total) leaving the surface that contains net charge denoted by Q is

total = Q0    …………………(1)

The permittivity of free space is

0 = 8.85410-12N-1C²m-2

The charge of the sphere will be

Q = 1.44710-3C

When we substitute these in (1), we get,

total = 1.44710-38.85410-12

∴total = 1.6310-8NC-1m²

Hence, the total electric flux that will be leaving the sphere’s surface will be 1.6310-8NC-1m².

Short Questions and Answers (5 Marks)  

Q.1. answer the following questions..

(a) Define equipotential surface. Demonstrate that an equipotential surface has an electric field that is always perpendicular to it.

(b) Find an expression for the potential at a location along a short electric dipole’s axial line.

Ans (a).  A surface with uniform potential is known as an equipotential surface.

We know that

dW = (qE)dx

(F = (qE), force on the test charge)

Since there is no work done when moving a test charge across an equipotential surface,

(b) Consider an electric dipole with a dipole length of 2a and a point P on the axial line where O, the dipole’s centre, is such that OP = r.

V=K(-q)(r+a)+K(+q)(r-a)

V=Kq1r-a-!r+a

V=Kqr+a-r+a(r-a)(r+a)

V = Kq2ar2-a2              (∴P = 2aq)

V = KPr2-a2

If a dipole has a short length, “a” can be neglected

Q.2. Check whether the ratio K e 2 G m e m p will be dimensionless. You can take the help of the physical constant’s table and check the value of this ratio. What is the significance of the ratio?

Ans. The given ratio is Ke2Gmemp.

Here, G is the gravitational constant and is denoted by Nm²kg-2.

me  is the mass of the electron, and mp is the mass of a proton. kge is the electric charge and is denoted by C. permittivity of free space is, 0 and its unit is Nm²C-2.

Hence the unit of the ratio Ke2Gmemp is

= (Nm2C-2)(C2)(Nm²kg-2)(kg)(kg)

The dimension can be given as = (M0L0T0)

e = 1.610-19C

G = 6.6710-11Nm²kg-2

me= 9.110-31kg

mp = 1.6610-27 kg,

Therefore, the numerical value of the ratio will be

Ke2Gmemp = 9109(1.610-19)26.6710-119.110-311.6610-27≈2.31039

This ratio illustrates the relationship between the gravitational and electric forces acting on a proton and an electron while maintaining their relative distance from one another.

Q.3. Four charges named q A =2 C, q B =-5 C, q C =2 C, and q D =-5 C are situated in the corners of a 10 cm square ABCD. What is the force acting on a 1 C charge in the square’s centre?

Ans. Four charges are positioned at the corners of the square in the given figure, which has four sides that are each 10 cm long. The square’s centre is O.

AB, BC, CD, and AD are the sides of the square. The diagonals of the square of length 102cm, AC and BD, are each 10 cm long.

AO, OB, OC, OD, are the length of 52cm.

The charge of the amount 1C is located at the centre of the square.

The sides of the square are AB, BC, CD, and AD. The diagonals of the square with a length of 10 cm each are AC and BD.

When compared to the repulsion between the charges placed at C and O, the force between the charges positioned at A and O is equally strong, but is directed in the other direction. They will therefore balance each other out.

Similar to the attraction force between the charges placed at D and O, there is an attraction force between charges placed at B and O that is similar in magnitude but moves in the opposite direction. As a result, their forces cancel one another out.

As a result, the 1C charge located at centre O is not subject to any net force from the four charges placed at the corners of the square.

Q.4. A system having two charges q A =2.5 10 -7 and q B =-2.5 10 -7 C is positioned at A(0,0,-15) and B(0,0,15). What will the electric dipole moment and total charge of the system be?

The value of charge at A, qA = 2.510-7

The value of charge at B, qB = -2.510-7C

Net amount of charge, qnet = qA+qB

qnet = +2.510-7 -2.510-7C

The distance between two charges at A and B,

d = 15+15 = 30cm

The electric dipole moment of the system can be represented by

P = qAd = qB0.3

P = 7.510-8 Cm along the z-axis.

Hence, the electric dipole moment of the system will be 7.510-8, and it will be along the positive z-axis.

Q.5. (a) The centres of two insulated charged copper spheres, A and B are separated by a distance of 50 cm. If the charge on each is 6.5 10 -7 , what is the electrostatic repulsion’s mutual force? The separation distance is much greater than the radii of A and B.

(b) if each sphere is charged twice as much as above and their distance is halved, what is the force of repulsion.

Ans. (a) Charge on the sphere, A,qA = charge on sphere B,qB = 6.510-7

The distance between the sphere, r = 50cm = 0.5m

The force of repulsion between the two spheres,

F = qAqB40r2

0 = free space permittivity

140 = 9109 Nm²C-2

∴F = 9109(6.510-7)2(0.5)2

= 1.5210-2N

Hence, the force that would exist between the two spheres is

When the charge is doubled, charge on the sphere

A,qA= charge on the sphere

B,qB= 26.510-7C=1.310-6C.

The distance that would remain between the spheres will be halved.

∴r = 0.52 = 0.25m

Force of repulsion that would exist between the two spheres

= 91091.310-51.310-5(0.25)2

= 161.5210-2

Hence, 0.243 Ns will be the force that would exist between the spheres.

Q.6. The entire electric field in a particular area of space is along the z-axis. Though it is not constant, the strength of the electric field grows uniformly along the positive z-direction at a rate of 105 NC -1 per metre. What torque and force does a system with a total dipole moment of 10 -7 Cm in the negative z-direction experience?

Ans. The dipole moment of the system, p = qdl = -10-7

The rate at which the electric field per unit increases

dedl = 10-5 NC-1

Force (f) that the system experiences can be described by the following relation,

F = qdEDldl

= -10-710-5

Hence, -10-2N will be the force in the negative z-direction or the opposite to the direction of the electric field. Therefore, the angle that would exist between the dipole moment and the electric field is 180°.

Torque(T) can be described by the relation,

(T) = pE sin180° = 0

Hence, the torque that the system will experience will be zero.

Q.7. (a) Take a random electrostatic field arrangement as an example. A small test charge is positioned at the configuration’s null point or the location where E = 0.

Show that the test charge’s equilibrium is inherently unstable., (b) verify this result using the basic arrangement of two charges with the same magnitude and sign spaced apart..

Ans. (a) Let’s start by assuming that the little test charge positioned at the null point of

the setup is stable and in equilibrium.

In order for the equilibrium to remain stable, the test charge must only move very slightly.

The charge will return to the null point in any direction because there will be surrounding it are tremendous restoring powers.

This shows even more strongly that all electric lines of force acting near the null point are toward the specified null point and inward.

However, according to Gauss’s law, the net electric flow through a chargeless surface enclosed is equal to zero. This fact defies the presumption that we had commenced with. In light of this, it may be said that the balance of the test charge must be unstable.

(b) The null point occurs to be at the midpoint of the line connecting these two charges when we take into account this configuration setup with two charges of the same sign and magnitude arranged at a specific distance apart.

According to the presumption made before, when the test charge is positioned at this midpoint, it will encounter substantial restoring forces as it tries to move itself.

However, because there is no restoring force along the normal to the line taken into account, the test charge is pulled off when it tries to displace in a direction normal to the line connecting the two charges.

The assumption is further challenged in this situation, since stable equilibrium prioritizes restoring force in all directions.

Important Question of Physics Class 12 Chapter 1

Important questions for class 12 physics chapter 1 electric charges and fields.

When you practice these Class 12 Physics Chapter 1 Important Questions, you will learn many things that will help you enhance your fundamental knowledge and be able to obtain good marks in the exams. Class 12 Physics Chapter 1 Important Questions are a lifesaver for those students who have a weak foundation of these concepts. Some of the knowledge that students learn during studying physics class 12 ch 1 key questions are as follows.

Electric Charge

The term ‘electricity’ is derived from the Greek word elektron, which means amber. The electric and magnetic forces existing inside matter, atoms, and molecules are employed to establish their properties. There are just two forms of the entity, which are known as electric charge.

An experiment was carried out to examine the electric charge. This experiment indicated that there exist two kinds of electrification, which are: (i) like charges repel from one another and (ii) unlike charges seem to attract one another. There’s a property which separates these two sorts of charges known as the polarity of charge.

Electric charge is recognized as the physical feature of matter which allows it to experience a force whenever it gets put on an electromagnetic field. Electric charges are separated into two sorts which are positive and negative. The negatively charged matter is called an electron, whereas a positively charged matter is called a proton.

Properties of Electric Charge

Some fundamental features of the electric charge are as follows.

Additive: This feature of electric charge implies that the total charge of a body can represent the complete sum of all the individual charges that appear to be working on the system.

Additive: This feature of electric charge implies that the total charge of a body can represent the complete sum of all the individual charges that appear to be working on the system. Quantization: This feature of electric charge implies that the total charge of a body can reflect the integral multiple of a basic quantum of charge.

Conservation: This feature of electric charge implies that the total or the full charge of a given system remains unchanged with time. In simple words, when an item gets charged because of some friction, a transition of charge from one item to another occurs. A charge is something that can’t be destroyed or created.

Electric Field

An electric field is described as an electric force present in one unit of charge. The field’s direction is believed to be identical to the direction of the force and assumed to exert a positive test charge. When the electric field is radially outward, then it is regarded as a positive charge, and when the electric field is radially inward, then it is regarded as a negative charge. Volt per metre is the SI unit of electric fields.

We may describe the electric field as a physical field which can be found encircling each electric charge and responsible for exerting force on all the other charges existing on the field. Electric fields are considered to be produced from time-varying or electric magnetic fields. Both electric fields and magnetic fields are viewed as representations of electromagnetic force, which is among the four fundamental forces that are present in the environment or nature.

Properties of Electric Field

Given below are some properties of an electric field.

• The lines present in an electric field never intersect with one another.
• The surface charge is perpendicular to the lines in the electric field.
• When the lines are close to each other, the field is strong, but when the lines are free and are not near together, the field is weak.
• The total lines in the field are exactly proportional to the amount of the charge.
• The lines present in the electric field seem to originate from a positive charge and finish with a negative charge.
• When it comes to a single charge, they seem to start and stop at infinity.
• The line curves of the electric field are continuously in a charge-free zone.

These are the basic theories of the chapter ‘Electric Charges and Fields’. This chapter contains more advanced concepts and theories that students will learn when they go through this chapter. It is advisable for students to practise the Class 12 Physics Chapter 1 Important Questions regularly to learn the advanced concepts so that they will be able to achieve good marks in the exams.

Class 12 Physics Chapter 1 Important Questions

Some questions that may come up in your Physics exams are as follows.

• What do you mean by the electric charge? Explain the team using an experiment.
• What do you mean by the term conductors?
• What do you mean by the term insulators?
• Explain the difference between the conductors and insulators.
• What is Coulomb’s law? Explain it with some examples and experiments.
• What is an electric dipole?

Benefits of Important Questions of Physics Class 12 Chapter 1

These Class 12 Physics Chapter 1 Important Questions will be advantageous for students who are weak in science. These notes will act as a guide for every pupil. Some of the advantages of referring to crucial questions for Class 12 Physics Chapter 1 are as follows.

• The questions which are provided are selected while considering the format and the CBSE syllabus as prescribed because students can lose a lot of marks if those norms are not followed.
• These questions are selected by subject experts.
• The questions are prepared while keeping in mind the intellectual aptitude of students so that they can understand every question.
• The questions that are mentioned in these notes are most likely to appear in the final exams, therefore prepping the students better.

It would be beneficial for students to refer to these Class 12 Physics Chapter 1 Important Questions so that they can understand the concepts better and would be able to achieve good marks.

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FAQs (Frequently Asked Questions)

1. which direction is a dipole positioned in a constant electric field to achieve the following a stable equilibrium b unstable equilibrium.

(a) A dipole is positioned parallel to the electric field to achieve stable equilibrium.

(b) A dipole is positioned antiparallel to the electric field for an unstable equilibrium.

2. Why must the electrostatic field at every point of a charged conductor be normal to the surface?

The direction of the electric field at that location is indicated by the tangent on a charged conductor.

3. In the middle of a cube with side l is a charge designated q, What is the electric flux going through the cube two opposed faces?

4. is the electric field caused by a combination of charges where the total charge is zero necessarily zero justify..

No, it isn’t always zero. If an electric dipole’s electric field is not zero, the electric field resulting from a charge arrangement with total charge is not zero.

5. What is the intensity of the electric field?

The electric field is the force on a unit positive (test) charge. Electric field = Force/Change

6. What is electrostatic shielding?

Electrostatic shielding is a phenomenon where an individual or an object is protected from an electric field or from an electric current by being housed inside a hollow conductor.

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Case Study Questions Class 12 Physics Electric Charges and Fields

Case study questions class 12 physics chapter 1 electric charges and fields.

CBSE Class 12 Case Study Questions Physics Electric Charges and Fields. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Electric Charges and Fields.

1.) Read the following paragraph and answer the following questions from 1.1 to  1.4 :

Charge is a property of matter that can be in two forms: positive charge and negative charge. If two objects which are having the same sign of charge repel to each other,  and on the other hand objects which have opposite sign charges attract to each other. Electric field concept comes from

Electric field is a vector quantity that means it has magnitude and direction both.

1.2   Choose the correct options from the following given below.

1.3   If Force acting on a test charge of 5C , is 25 N along north direction then what will be electric field intensity at that point.

When both bodies have the same charges they will repel each other and on the other hand they will attract each other when both bodies have different charges.

So E = F/q = 25/5 = 5 N/C

2.) Read the following paragraph and answer the following questions from 2.1 to 2.3

As we know electric field is a vector quantity so due to more than one point charges electric field intensity can be calculated at any point as a vector sum of due to all point charges intensity and direction will be along the resultant vector direction.If at any point electric field is uniform can be drawn as a parallel line and if electric field is non- uniform it can be drawn non-parallel line but when then electric field line near to each other we can say there is a strong electric field compare to other point, where electric field line is little bit far away that means we can say electric field line density is directly proportional to electric field strength.And this electric field lines never intersect to each other. The electric field strength measure in V/m

And It’s SI unit is defined is a Volt/meter

3.) Read the following paragraph and answer the following questions from 3.1 to 3.3

Assertion(A): Electric force between two point charges depends upon the magnitudes of the two charges and  also depends upon the distance between them.

C. Assertion is true, but Reason is false.

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CBSE 12th Standard Physics Subject Electric Charges And Fields Chapter Case Study Questions 2021

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 12 , and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams 

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Cbse 12th standard physics subject electric charges and fields case study questions 2021.

12th Standard CBSE

Final Semester - June 2015

(ii) Which one of the following charges is possible?

(iii) If a charge on a body is 1 nC, then how many electrons are present on the body?

(iv) If a body gives out 10 9 electrons every second, how much time is required to get a total charge of 1 C from it?

(v) A polythene piece rubbed with wool is found to have a negative charge of3.2 x 1O -7 C.Calculate the number of electrons transferred.

(ii) Extra electrons on this particular oil drop (given the presently known charge of the electron) are

(iii) A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field 100 V m -1 .If the mass of the drop is 1.6 X 10 -3 g, the number of electrons carried by the drop is (g= 10 m s -2 )

(iv) The important conclusion given by Millikan's experiment about the charge is

(v) If in Millikan's oil drop experiment, charges on drops are found to be  \(8 \mu \mathrm{C}, 12 \mu \mathrm{C}, 20 \mu \mathrm{C}\)  then quanta of charge is

(ii) Which of the following is false for electric lines of force?

(iii) Which one of the following pattern of electric line of force in not possible in filed due to stationary charges?

(iv) Electric lines of force are curved

(ii) An electric charge of 8.85 X 10 -13 C is placed at the centre of a sphere of radius 1 m. The electric flux through the sphere is

(iv) What charge would be required to electrify a sphere of radius 25 cm so as to get a surface charge density of  \(\frac{3}{\pi} \mathrm{C} \mathrm{m}^{-2} ?\)

(v) The SI unit of linear charge density is

(ii) E in the outer region of the second plate is

(iii) E between the plates is

(iv) The ratio of E from right side of B at distances 2 cm and 4 ern, respectively is

(v) In order to estimate the electric field due to a thin finite plane metal plate, the Gaussian surface considered is

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Cbse 12th standard physics subject electric charges and fields case study questions 2021 answer keys.

(i) (d) (ii) (b):  From,  \(q=n e, n=\frac{q}{e}=\frac{3.2 \times 10^{-18}}{1.6 \times 10^{-19}}=20\) As n is an integer, hence this value of charge is possible. (iii) (d): Charge on the body is q = ne   \(\therefore\)  No. of electrons present on the body is \(n=\frac{q}{e}=\frac{1 \times 10^{-9} \mathrm{C}}{1.6 \times 10^{-19} \mathrm{C}}=6.25 \times 10^{9}\) (iv) (C): Here, n = 10 9 electrons per second, Charge given per second,  q = ne = 10 9 x 1.6 x 10 -19 C q = 1.6 X 10 -10 C Total charge, Q = 1 C (given) \(\therefore\)   Time required  \(=\frac{Q}{q}=\frac{1}{1.6 \times 10^{-10}} \mathrm{~s}=6.25 \times 10^{9} \mathrm{~s}\) \(\therefore\)   \(\frac{6.25 \times 10^{9}}{3600 \times 24 \times 365} \text { year }=198.19 \text { years. }\) (v) (a):  As q = ne, n =  \(\frac{3.2 \times 10^{-7}}{1.6 \times 10^{-19}}\) ⇒ n = 2 x 10 12 electrons.

(i) (a):  As,  \(q E=m g \Rightarrow q=\frac{1.08 \times 10^{-14} \times 9.8}{1.68 \times 10^{5}}\) \(=6.4 \times 10^{-19} \mathrm{C}\) (ii) (a):  \(q=n e \text { or } \Rightarrow n=\frac{6.4 \times 10^{-19}}{1.6 \times 10^{-19}}=4\) (iii) (c) : For the drop to be stationary, Force on the drop due to electric field = Weight of the drop qE=mg \(q=\frac{m g}{E}=\frac{1.6 \times 10^{-6} \times 10}{100}=1.6 \times 10^{-7} \mathrm{C}\) Number of electrons carried by the drop is \(n=\frac{q}{e}=\frac{1.6 \times 10^{-7} \mathrm{C}}{1.6 \times 10^{-19} \mathrm{C}}=10^{12}\) (iv) (c) (v) (d): Millikan's experiment confirmed that the charges are quantized, i.e., charges are small integer multiples of the base value which is charge on electron. The charges on the drops are found to be multiple of 4. Hence, the quanta of charge is 4 \(\mu \) C.

(i) (a)  radially outwards (ii) (c): Electric lines offorce do not form any closed loops. (iii) (c) : Electric field lines can't be closed. (iv) (d)  (d) both (b) and (c) (v) (a)  \(\text { (a) } E_{A}>E_{B}>E_{C}\)

(i) (a): Gauss's law is applicable for any closed surface. Gauss's law is most useful in situation where the charge distribution has spherical or cylindrical symmetry or is distributed uniformly over the plane. Whereas electric dipole is a system of two equal and opposite point charges separated by a very small and finite distance. So both statements are correct. (ii) (b): According to Gauss's law, the electric flux through the sphere is  \(\phi=\frac{q_{\mathrm{in}}}{\varepsilon_{0}}=\frac{8.85 \times 10^{-13} \mathrm{C}}{8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}}=0.1 \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^{2}\) (iii) (c) : For uniformly volume charge density, \(E=\frac{\rho r}{3 \varepsilon_{0}}\) \(E \propto r\) (iv) (a) : r = 25 ern = 0.25 m  \(\sigma=\frac{3}{\pi} \mathrm{C} / \mathrm{m}^{2}\) As,  \(\sigma=\frac{q}{4 \pi r^{2}} \Rightarrow q=4 \pi \times(0.25)^{2} \times \frac{3}{\pi}=0.75 \mathrm{C}\) (v) (b): The line charge density at a point on a line is the charge per unit length of the line at that point \(\lambda=\frac{d q}{d L}\) Thus, the SI unit for  \(\lambda \text { is } \mathrm{Cm}^{-1} \text {. }\)

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Case study question for class 12 physics chapter 6 electromagnetic induction, case study questions for class 12 physics chapter 5 magnetism and matter, case study questions for class 12 physics chapter 10 wave optics, case study questions for class 12 physics chapter 8 electromagnetic waves, case study questions for class 12 physics chapter 14 semiconductor electronics, case study question for class 12 physics chapter 4 moving charges and magnetism, case study questions for class 12 physics chapter 2 electrostatic potential and capacitance, case study questions for class 12 physics chapter 13 nuclei, case study questions for class 12 physics chapter 12 atoms, case study questions for class 12 physics chapter 11 dual nature of radiation and matter, case study questions for class 12 physics chapter 9 ray optics and optical instruments, case study question for class 12 physics chapter 3 current electricity, case study question for class 12 physics chapter 1 electric charges and fields.

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NCERT Solutions for Class 12 Physics Chapter 1 PDF Download

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NCERT Solutions for Class 12 Physics Chapter 1 PDF

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NCERT Solutions for Class 12 Physics Chapter 1 Free PDF Download

Ncert solutions for class 12 physics chapter 1 electric charges and fields.

This article deals with NCERT Solutions for Class 12 Physics Chapter 1. Physics is certainly a complicated branch of science. However, NCERT Solutions makes this subject pretty easy. Most noteworthy, these solutions clear all fundamental concepts. Hence, the details of Physics become crystal clear for those who learn from NCERT Solutions

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CBSE Class 12 Physics Chapter 1 – Electric Charges and Fields NCERT Solutions

NCERT Solutions for Class 12 Physics Chapter 1 deals with Electric Charges and Fields. This chapter deals with electricity, charge, and discharge. Moreover, it deals with the generation of static electricity. There is certainly a discussion of the various components of electrostatics here.

Subtopics covered under NCERT Solutions for Class 12 Physics Chapter 1

1.1 Introduction

1.2 Electric Charge

1.3 Conductors and Insulators

1.4 Charging by Induction

1.5 Basic Properties of Electric Charge

• 1.5.1 Additivity of charges
• 1.5.2 Charge is conserved
• 1.5.3 Quantisation of charge

1.6 Coulomb’s Law

1.7 Forces between Multiple Charges

1.8 Electric Field

• 1.8.1 Electric field due to a system of charges
• 1.8.2 Physical significance of electric field

1.9 Electric Field Lines

1.10 Electric Flux

1.11Electric Dipole

• 1.11.1 The field of an electric dipole
• 1.11.2 Physical significance of dipoles

1.12 Dipole in a Uniform External Field

1.13 Continuous Charge Distribution

1.14 Gauss’s Law

1.15 Applications of Gauss’s Law

• 1.15.1 Field due to an infinitely long straight uniformly charged wire
• 1.15.2 Field due to a uniformly charged infinite plane sheet
• 1.15.3 Field due to a uniformly charged thin spherical shell

NCERT Solutions for Class 12 Physics Chapter 1

First of all, this is a vital part of Physics. Moreover, it has applications in a number of fields of science. Furthermore, the chapter will help students in becoming engineers and architects. This is due to the huge application of electricity and related concepts in these fields.

Below is a brief explanation of the sub-units of chapter 1:

1.1 Introduction –

First of all, this part gives a general idea of Electric Charges and Fields. This sub-unit certainly gives a general introduction to the chapter.

1.2 Electric Charge –

This refers to the physical property of matter that causes it to experience a force. Moreover, it happens when the matter is placed in an electromagnetic field.

1.3 Conductors and Insulators –

Conductors are substances which allow electricity to pass through them easily. In contrast, insulators are those which offer high resistance to the passage of electricity.

1.4 Charging by Induction –

This is certainly a process. Most noteworthy, in this process, metal spheres will each be equal and oppositely charged.

1.5 Basic Properties of Electric Charge –

This part certainly discusses the basic properties of electric charge. Furthermore, the minor units here are:

1.6 Coulomb’s Law –

This refers to a quantitative statement about the force between two point charges.

1.7 Forces between Multiple Charges –

This part provides information on the forces between multiple charges.

1.8 Electric Field –

This can be defined as the electric force per unit charge. The minor parts here are:

1.9 Electric Field Lines –

Electric field lines certainly provide a means to visualize the electric field.

1.10 Electric Flux –

This is an analogous quantity in the case of the electric field.

1.11Electric Dipole –

This refers to a pair of equal and opposite point charges q and –q. Furthermore, this is separated by a distance 2a. The minor units here are below:

1.12 Dipole in a Uniform External Field –

This part certainly deals with a dipole in a uniform external field.

1.13 Continuous Charge Distribution –

The continuous charge distribution is pretty similar to that we adopt for continuous mass distribution.

1.14 Gauss’s Law –

This law tells that the total electric flux through a closed surface is zero. Furthermore, it happens if no charge is enclosed by the surface.

1.15 Applications of Gauss’s Law –

This part certainly contains applications of Gauss’s Law. It has the following parts:

You can download NCERT Solutions for Class 12 Physics Chapter 1 PDF by clicking on the button below.

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Solved Questions For You:

Question 1.  There two types of electric charges positive charges and negative charges. The property which differentiates the two types of charges is:

Question 2. What will happen when we rub a glass rod with silk cloth?

Question 3. When a person combs his hair, static electricity is generated sometimes by what process?

Question 4.  A method for charging a conductor without bringing a charged body in contact with it is called:

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MCQ Questions for Class 12 Physics Chapter 1 Electric Charges and Fields with Answers

We have compiled the NCERT MCQ Questions for Class 12 Physics Chapter 1 Electric Charges and Fields with Answers Pdf free download covering the entire syllabus. Practice MCQ Questions for Class 12 Physics with Answers on a daily basis and score well in exams. Refer to the Electric Charges and Fields Class 12 MCQs Questions with Answers here along with a detailed explanation.

Electric Charges and Fields Class 12 MCQs Questions with Answers

Question 1. In comparison With the electrostatic force between two electrons, the electrostatic force between two protons is: (a) greater (b) smaller (c) zero (d) same.

Answer: (d) same.

Question 2. If electric flux through a closed surface is zero. It means that: (a) the net charge inside the surface is zero. (b) the electric field is necessarily zero at all points on the surface. (c) no charge exists inside the surface. (d) no charge exists outside the surface.

Answer: (a) the net charge inside the surface is zero.

Question 3. Static electricity is produced by (a) induction only (b) friction (c) chemical reaction only (d) friction and induction both.

Answer: (b) friction

Question 4. When distance between two charges is reduced to one-half of the original distance, the force between them will remain the same if one of the charges is made: (a) one fourth (b) four times (c) double (d) one-halftimes.

Answer: (a) one fourth

Question 5. A unit Coulomb charge is one which when placed in air at a distance of 1 m from an equal and similar charge repel it with a force of (a) 9 × 10 9 N (b) 1 N (c) 1 dyne (d) None of these.

Answer: (a) 9 × 10 9 N

Question 6. When two charged spheres are connected with a wire, the electric charge on them is shared: (a) inversely as their capacity (b) equally (c) in proportional to their capacity (d) None of these.

Answer: (b) equally

Question 7. An electric dipole of moment p is placed in the position of stable equilibrium in a uniform electric field \(\vec{E}\). The couple required to rotate it through an angle 6 the initial position is: (a) -PE cos θ (b) PE tan θ (c) PE cos θ (d) PE sin θ.

Answer: (d) PE sin θ.

Question 8. Two point charges each of 20 µC are placed 50 cm apart in air. What is the electric field intensity at the mid point on the line joining the centre of two point charges? (a) 5 × 10 6 NC -1 (b) 18 × 10 6 NC -1 (c) Zero (d) None of these

Answer: (c) Zero

Question 9. Two charges 10 pC and 5 pC are placed 20 cm apart. The ratio of Coulomb’s force experienced by there is: (a) 2 : 5 (b) 1 : 1 (c) √3 = √7 (d) None of these

Answer: (b) 1 : 1

Answer: (a) E 1 > E 2

Question 11. The surface charge density oil the copper sphere is σ. The electric field strength on the surface of Sphere of radius r is: (a) \(\frac {σ}{2}\) (b) σ (c) \(\frac {σ}{2ε_0}\) (d) σ/ε 0

Answer: (d) σ/ε 0

Answer: (b) 8 electrons can be removed

Question 13. Which of the following method is an indirect method of charging: (a) charging by induction (b) charging by physical contact (c) charging by friction (d) none of these.

Answer: (a) charging by induction

Question 14. An electron and a proton are placed in the same uniform electric field. What will be the ratio of the acceleration of electron to that of proton? (a) 1 (b) zero (c) \(\frac {m_p}{m_e}\) (d) \(\frac {m_e}{m_p}\)

Answer: (c) \(\frac {m_p}{m_e}\)

Question 15. The position of the charge inside the enclosing surface is changed in such a way that the total charge remains constant. Then the total normal electric flux through the enclosing surface: (a) increases (b) decreases (c) changes erratically (d) Remains unchanged.

Answer: (d) Remains unchanged.

Fill in the Blanks

Question 1. The force between two point charges ………………… when dielectric constant of the medium in which they are held increases.

Answer: decreases.

Question 2. When a dielectric medium is placed in a uniform electric field, the dielectric constant of the medium is the ratio of the ……………….. of the medium to the ……………… of free space.

Answer: absolute permittivity, absolute permittivity.

Answer: Minimum.

Question 4. The number of electric lines of force that radiate outward from 1C charge is ………………

Answer: 1.13 × 10 11

Question 5. ……………… is a central force in nature.

Answer: Electrostatic force.

Hope the information shed above regarding NCERT MCQ Questions for Class 12 Physics Chapter 1 Electric Charges and Fields with Answers Pdf free download has been useful to an extent. If you have any other queries of CBSE Class 12 Physics Electric Charges and Fields MCQs Multiple Choice Questions with Answers, feel free to reach us so that we can revert back to us at the earliest possible.