NCERT Exemplar for Class 12 Physics Chapter 1 physics class 12 electric charges and fields

Physics is a tricky yet interesting subject which opens the door for various new theories and experiments for students. To keep the curiosity awakened in students and understand the concepts more closely, students must solve the NCERT exemplar for Class 12 Physics. Do not worry, if you are stuck between solving the questions of the chapters, SimplyAcad has brought you the NCERT Exemplar for Class 12 Physics Chapter 1 Electric Charges and Fields below to help students learn about the different topics in a detailed manner. The exemplar will allow students to gain deep insights of all the sections and prepares you better for the upcoming 12th board examinations.

The given exemplar contains MCQs of two different types, very short and long answer Type Questions, there are a total of 31 questions asked. Students can access the NCERT exemplar for class 12 physics, Chapter 1: Solid State by scrolling below. Along with this, there are several NCERT exemplar for class 12 science of all the chapters provided on this platform.

Access the NCERT Exemplar Class 12 Physics Chapter 1 Electric Charges and Fields

Multiple Choice Questions I for physics class 12 electric charges and fields

1. In the figure, two positive charges, \( q_2 \) and \( q_3 \), are fixed along the y-axis and exert a net electric force in the \( +x \) direction on a charge \( q_1 \) fixed along the x-axis. If a positive charge \( Q \) is added at \( (x, 0) \), the force on \( q_1 \): \begin{itemize} \item[A)] Shall increase along the positive x-axis \item[B)] Shall decrease along the positive x-axis \item[C)] Shall point along the negative x-axis \item[D)] Shall increase but the direction changes because of the intersection of \( Q \) with \( q_1 \) and \( q_3 \) \end{itemize} \textbf

{Solution:} \\ Since positive charges \( q_2 \) and \( q_3 \) exert a net force in the \( +x \) direction on the charge \( q_1 \) fixed along the x-axis, the charge \( q_1 \) is negative. Adding a positive charge \( Q \) at \( (x, 0) \) will exert an additional repulsive force on the negative charge \( q_1 \). Thus, the force on \( q_1 \) will increase along the positive x-axis due to the repulsive force from \( Q \). The correct option is: \textbf{A) Shall increase along the positive x-axis}

1.2. A point positive charge is brought near an isolated conducting sphere. The electric field is best given by: \begin{itemize} \item[A)] Figure (i) \item[B)] Figure (ii) \item[C)] Figure (iii) \item[D)] Figure (iv) \end{itemize} \textbf

{Solution:} \\ The correct option is A. Figure (i). \\ When a point positive charge is brought near an isolated conducting sphere, the following occurs: \begin{itemize} \item Some negative charge develops on the side of the sphere closest to the positive point charge. \item An equal amount of positive charge develops on the far side of the sphere. \end{itemize} The electric field lines of force emanating from the point positive charge end normally on the left side of the sphere (where the negative charge is induced). Additionally, due to the positive charge on the right side of the sphere, the electric field lines will emanate normally from the right side of the sphere. Thus, the electric field configuration is best represented by Figure (i).

1.3. The electric flux through the surface in the following figures: \begin{itemize} \item[A)] in Fig. (iii) is the least \item[B)] in Fig. (iv) is the largest \item[C)] is the same for all the figures \item[D)] in Fig. (ii) is the same as Fig. (iii) but is smaller than Fig. (iv) \end{itemize} \textbf

{Solution:} \\ The correct option is C. \\ Electric flux (\(\Phi\)) is defined as the number of electric field lines passing through a given cross-sectional area. According to Gauss’s Law: \[ \Phi = \oint \vec{E} \cdot d\vec{A} = \frac{q}{\epsilon_0} \] where \( \vec{E} \) is the electric field, \( d\vec{A} \) is the differential area vector, \( q \) is the charge enclosed by the surface, and \( \epsilon_0 \) is the permittivity of free space. Electric flux through a surface depends on the amount of charge enclosed by the surface and is independent of the shape, size, or area of the surface. Given that the charge enclosed by the surfaces in all figures is the same, the electric flux through all the surfaces should be the same. Thus, the correct answer is: \textbf{C) is the same for all the figures.}

1.4. Five charges \( q_1, q_2, q_3, q_4 \), and \( q_5 \) are fixed at their positions as shown in the figure. \( S \) is a Gaussian surface. The Gauss’s law given by the following equation: \[ \oint \vec{E} \cdot d\vec{A} = \frac{q}{\epsilon_0} \] Which of the following statements is correct? \begin{itemize} \item[A)] \( \vec{E} \) on the LHS of the above equation will have a contribution from \( q_1, q_5 \), and \( q_3 \) while \( q \) on the RHS will have a contribution from \( q_2 \) and \( q_4 \) only. \item[B)] \( \vec{E} \) on the LHS of the above equation will have a contribution from all charges while \( q \) on the RHS will have a contribution from \( q_2 \) and \( q_4 \) only. \item[C)] \( \vec{E} \) on the LHS of the above equation will have a contribution from all charges while \( q \) on the RHS will have a contribution from \( q_1, q_3 \), and \( q_5 \) only. \item[D)] Both \( \vec{E} \) on the LHS and \( q \) on the RHS will have contributions from \( q_2 \) and \( q_4 \) only. \end{itemize} \textbf

{Solution:} \\ The correct option is B. \\ According to Gauss’s Law, the electric flux through a closed surface depends only on the charge enclosed by that surface. The electric field \( \vec{E} \) on the LHS of the equation is due to all charges present both inside and outside the Gaussian surface. However, the electric flux, represented by \( \oint \vec{E} \cdot d\vec{A} \), depends only on the charges enclosed by the surface. Therefore, \( \vec{E} \) on the LHS will have a contribution from all charges, whereas \( q \) on the RHS, which represents the total charge enclosed by the Gaussian surface, will have contributions from only \( q_2 \) and \( q_4 \), assuming \( q_2 \) and \( q_4 \) are the charges enclosed by the surface. Hence, the correct answer is: \textbf{B) \( \vec{E} \) on the LHS of the above equation will have a contribution from all charges while \( q \) on the RHS will have a contribution from \( q_2 \) and \( q_4 \) only.}

1.5. Figure shows electric field lines in which an electric dipole p is placed as shown. Which of the following statements is correct?

a) the dipole will not experience any force

b) the dipole will experience a force towards right

c) the dipole will experience a force towards left

d) the dipole will experience a force upwards

Answer:

The correct answer is c) the dipole will experience a force towards left

1.6. A point charge +q is placed at a distance d from an isolated conducting plane. The field at a point P on the other side of the plane is

a) directed perpendicular to the plane and away from the plane

b) directed perpendicularly to the plane towards the plane

c) directed radially away from the point charge

d) directed radially towards the point charge

Answer:

The correct answer is a) directed perpendicular to the plane and away from the plane

1.7. A hemisphere is uniformly charged positively. The electric field at a point on a diameter away from the centre is directed

a) perpendicular to the diameter

b) parallel to the diameter

c) at an angle tilted towards the diameter

d) at an angle tilted away from the diameter

Answer:

The correct answer is a) perpendicular to the diameter

Multiple Choice Questions II

1.8. If
over a surface, then

a) the electric field inside the surface and on it is zero

b) the electric field inside the surface is necessarily uniform

c) the number of flux lines entering the surface must be equal to the number of flux lines leaving it

d) all charges must necessarily be outside the surface

Answer:

The correct answers are

c) the number of flux lines entering the surface must be equal to the number of flux lines leaving it

d) all charges must necessarily be outside the surface

1.9. The electric field at a point is

a) always continuous

b) continuous if there is no charge at that point

c) discontinuous only if there is a negative charge at that point

d) discontinuous if there is a charge at that point

Answer:

The correct answers are

b) continuous if there is no charge at that point

d) discontinuous if there is a charge at that point

1.10. If there were only one type of charge in the universe, then

a) on any surface

b)if the charge is outside the surface

c)could not be defined

d)if charges of magnitude q were inside the surface

Answer:

The correct answers are

b) if the charge is outside the surface

d) if charges of magnitude q were inside the surface

1.11. Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region

a) the electric field is necessarily zero

b) the electric field is due to the dipole moment of the charge distribution only

c) the dominant electric field is proportional to 1/r3 for large r, where r is the distance from an origin in this region

d) the work done to move a charged particle along a closed path, away from the region, will be zero

Answer:

The correct answers are

c) the dominant electric field is proportional to 1/r3 for large r, where r is the distance from an origin in this region

d) the work done to move a charged particle along a closed path, away from the region, will be zero

1.12. Refer to the arrangement of charges in the figure and a Gaussian surface of radius \( R \) with \( Q \) at the center. Then: \begin{itemize} \item[A)] Total flux through the surface of the sphere is \( \frac{Q}{\epsilon_0} \). \item[B)] Field on the surface of the sphere is \( -\frac{Q}{4 \pi \epsilon_0 R^2} \). \item[C)] Field on the surface of the sphere due to \( -2Q \) is the same everywhere. \item[D)] Flux through the surface of the sphere due to \( 5Q \) is zero. \end{itemize} \textbf

{Solution:} \\ The correct option is D. \\ According to Gauss’s Law: \[ \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\epsilon_0} \] where \( q_{\text{enclosed}} \) is the charge enclosed by the Gaussian surface. Given: – Charge inside the Gaussian surface: \( Q \) and \( -2Q \) – Total charge enclosed by the Gaussian surface: \( Q – 2Q = -Q \) Therefore, the total flux through the surface of the sphere is: \[ \oint \vec{E} \cdot d\vec{A} = \frac{-Q}{\epsilon_0} \] The charge \( 5Q \) is outside the Gaussian surface and does not contribute to the flux through the surface. Thus, the flux through the surface of the sphere due to \( 5Q \) is zero. Hence, the correct answer is: \textbf{D) Flux through the surface of the sphere due to \( 5Q \) is zero.}

1.13. A positive charge Q is uniformly distributed along a circular ring of radius R. A small test charge q is placed at the centre of the ring. Then

a) if q > 0 and is displaced away from the centre in the plane of the ring., it will be pushed back towards the centre

b) if q < 0 and is displaced away from the centre in the plane of the ring, it will never return to the centre and will continue moving till it hits the ring

c) if q < 0, it will perform SHM for small displacement along the axis

d) q at the centre of the ring is in an unstable equilibrium within the plane of the ring for q > 0

Answer:

The correct answers are

a) if q > 0 and is displaced away from the centre in the plane of the ring., it will be pushed back towards the centre

b) if q < 0 and is displaced away from the centre in the plane of the ring, it will never return to the centre and will continue moving till it hits the ring

d) q at the centre of the ring is in an unstable equilibrium within the plane of the ring for q > 0

Very Short Answers

1.14. An arbitrary surface encloses a dipole. What is the electric flux through this surface? \textbf

{Solution:} \\ The net charge of a dipole is given by: \[ \text{Net charge} = -q + q = 0 \] According to Gauss’s theorem, the electric flux \(\Phi\) through a closed surface is given by: \[ \Phi = \frac{q_{\text{enclosed}}}{\epsilon_0} \] where \(q_{\text{enclosed}}\) is the total charge enclosed by the surface. For a dipole, the total enclosed charge is zero. Therefore: \[ \Phi = \frac{0}{\epsilon_0} = 0 \] Thus, the electric flux through the surface enclosing the dipole is: \textbf{0}

 

1.16. The dimensions of an atom are of the order of an Angstrom. Thus there must be large electric fields between the protons and electrons. Why, then is the electrostatic field inside a conductor zero?

Answer:

The electrostatic field inside a conductor is zero as the dimensions of the atom are in the order of an Angstrom and the electrostatic field inside the conductor is caused by the presence of the excess charges.

1.17. If the total charge enclosed by a surface is zero, does it imply that the electric field everywhere on the surface is zero? Conversely, if the electric field everywhere on a surface is zero, does it imply that net charge inside is zero.

Answer:

The total charge enclosed by a surface is zero, it doesn’t imply that the electric field everywhere on the surface is zero and the field may be normal to the surface. Also, the conversely it does imply that the electric field everywhere on the surface is zero.

1.18. Sketch the electric field lines for a uniformly charged hollow cylinder shown in the figure.

Answer:

Following are the side view and top view of the electric field lines for a uniformly charged hollow cylinder.

Short Answers for Physics Class 12 electric charges and fields

1.20. A paisa coin is made up of Al-Mg alloy and weighs 0.75 g. It has a square shape and its diagonal measures 17 mm. It is electrically neutral and contains equal amounts of positive and negative charges. Treating the paisa coins made up of only Al, find the magnitude of an equal number of positive and negative charges. What conclusion do you draw from this magnitude?

Answer:

1 molar mass M of Al = 6.023 × 1023 atoms

No.of aluminium atoms in one paise coin = N = 6.023 × 1023/26.9815 × 0.75 = 1.76 ×1022

There are 13 protons and 13 electrons on the Al as the charge number is 13

Magnitude of positive and negative charges = 3.48 × 104 C = 34.8 kC

1.21. Consider a coin of example 1.20. It is electrically neutral and contains equal amounts of positive and negative charge of magnitude 34.8 kC. Suppose that these equal charges were concentrated in two-point charges separated by

a) 1 cm

b) 100 m

Find the force on each such point charge in each of the three cases. What do you conclude from these results?

Answer:

a) F1 = |q|2/4πε0r12 = 1.09 × 1023 N

b) F2 = |q|2/4πε0r22 = 1.09 × 1015 N

c) F3 = |q|2/4πε0r32 = 1/09 × 107 N

1.22. Figure represents a crystal unit of caesium chloride, CsCl. The caesium atoms, represented by open circles are situated at the corners of a cube of side 0.40 nm, whereas a Cl atom is situated at the centre of the cube. The Cs atoms are deficient in one electron while the Cl atom carries an excess electron

a) what is the net electric field on the Cl atom due to eight Cs atoms?

b) suppose that the Cs atom at the corner A is missing. What is the net force now on the Cl atom due to seven remaining Cs atoms?

Answer:

a) The net electric field at the centre of the cube is zero as the chlorine atom is distributed equally from all the eight corners of the cube from the centre.

b) The vector sum of the electric field due to charge A is EA + Eseven charges = 0

 

1.24. Figure shows the electric field lines around three-point charges A, B, and C

a) which charges are positive?

b) which charge has the largest magnitude? Why?

c) in which region or regions of the picture could the electric field be zero?

i) near A

ii) near B

iii) near C

iv) nowhere

Answer:

a) A and C are positive as electric lines come out from the positive charge.

b) The magnitude of C is maximum than A and B because as the electric lines of forces from charge increases, the intensity of the electric field increases.

c) The electric field will be zero between A and C as that is the neutral point.

1.25. Five charges, q each are placed at the corners of a regular pentagon of side ‘a’

a) i) what will be the electric field at O, the centre of the pentagon?

ii) what will be the electric field at O if the charge from one of the corners is removed?

iii) what will be the electric field at O if the charge q at A is replaced by –q?

b) how would you answer to a) be affected if pentagon is replaced by an n-sided regular polygon with charge q at each of its corners?

Answer:

a) i) The net electric field at O will be zero as it is symmetric to all the five charges.

ii) Following is the electric field at O if the charge from one of the corners is removed:

E1 = -1q/4πε0r2

Also, the direction of the electric field will be from O to A.

iii) Following is the electric field at O if the charge q at A is replaced by –q:

E = -2q/4πε0r2

b) When an n-sided regular polygon with charge q is replaced at each of the corners, then

i) Again at centre O, the net electric field will be zero.

ii) The direction of the electric field will be from O to opposite of OA and the electric field is given as:

E2 = -q/4πε0r2

iii) The electric field will be from O to A and is given as:

E = -2q/4πε0r2

Long Answer

1.26. In 1959 Lyttleton and Bondi suggested that the expansion of the universe could be explained if matter carried a net charge. Suppose that the universe is made up of hydrogen atoms with a number density N, which is maintained a constant. Let the charge on the proton be: ep = -(1 + y) e where e is the electronic charge.

a) find the critical value of y such that expansion may start

b) show that the velocity of expansion is proportional to the distance from the centre

Answer:

a) There is one proton and one electron charge on each hydrogen atom.

eH = eP + e = -(1+y)e+e=-ye=|ye|

E is the electric field intensity at a distance R on the surface and according to Gauss theorem,

E = 1/3 N|ye|R/ε0

Solving for y we get, y = 10-18

b) Ne force experienced by the hydrogen atom is F = Fc – FG

The acceleration experienced by the hydrogen atom is given as d2R/dt2 = α2R

v = dR/dt = αR

Therefore, v is proportional to R that velocity is proportional to the distance from the centre.

1.31. 1) Total charge \( -Q \) is uniformly spread along the length of a ring of radius \( R \). A small test charge \( +q \) of mass \( m \) is kept at the center of the ring and is given a gentle push along the axis of the ring. \textbf{A) Show that the particle executes simple harmonic oscillation.} 2) Total charge \( -Q \) is uniformly spread along the length of a ring of radius \( R \). A small test charge \( +q \) of mass \( m \) is kept at the center of the ring and is given a gentle push along the axis of the ring. \textbf{B) Obtain its time period.} \textbf

{Solution:} \textbf{A)} The electric field \( E \) on the axis of a ring of radius \( R \) with total charge \( -Q \) is given by: \[ E = -\frac{kQz}{(R^2 + z^2)^{3/2}} \] where \( z \) is the distance along the axis from the center of the ring. The force \( F \) on the test charge \( q \) placed at a distance \( z \) from the center of the ring is: \[ F = qE = -\frac{kQqz}{(R^2 + z^2)^{3/2}} \] For small displacements where \( z \ll R \), we can approximate: \[ F \approx -\frac{kQqz}{R^3} \] This approximation shows that the force is proportional to \( -z \), indicating simple harmonic motion (SHM). The governing equation for SHM is: \[ F = -Kz \] where \( K = \frac{kQq}{R^3} \). \textbf{Final Answer:} \[ F = -\frac{kQqz}{R^3} \] \textbf{B)} To find the time period \( T \) of the oscillation, we use the formula for the time period of SHM: \[ T = 2\pi \sqrt{\frac{m}{K}} \] Substituting \( K = \frac{kQq}{R^3} \), we get: \[ T = 2\pi \sqrt{\frac{m}{\frac{kQq}{R^3}}} = 2\pi \sqrt{\frac{mR^3}{kQq}} \] Using \( k = \frac{1}{4\pi \epsilon_0} \), we have: \[ T = 2\pi \sqrt{\frac{mR^3}{\frac{1}{4\pi \epsilon_0} Qq}} = 2\pi \sqrt{\frac{4\pi \epsilon_0 m R^3}{Qq}} \] \textbf{Final Answer:} \[ T = 2\pi \sqrt{\frac{4\pi \epsilon_0 m R^3}{Qq}} \]

NCERT Exemplar For Class 12 Science

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