NCERT Solutions for Class 12 Physics Chapter 1

NCERT Solutions for Class 12 Physics Chapter 1

Q 1.1) What is the force between two small charged spheres having charges of 2 × 10–7C and 3 × 10–7 C placed 30 cm apart in the air?

Q 1.2) The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in the air is 0.2 N.

(a) What is the distance between the two spheres?

(b) What is the force on the second sphere due to the first?

Q 1.3) Check that the ratio ke2/G memp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Q 1.4) (i) Explain the meaning of the statement ‘electric charge of a body is quantised’.
(ii) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

Soln.:

(i) The ‘electric charge of a body is quantized’ means that only integral (1, 2, …n)  numbers of electrons can be transferred from a body to another.

Charges cannot get transferred in fractions. Hence, the total charge possessed by a body is only in integral multiples of electric charge.

(ii) In the case of large scale or macroscopic charges, the charge which is used over there is comparatively too huge to the magnitude of the electric charge. Hence, on a macroscopic level, the quantization of charge is of no use Therefore, it is ignored and the electric charge is considered to be continuous.

Q 1.5) 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.

Soln.:

When two bodies are rubbed against each other,  a charge is developed on both bodies. These charges are equal but opposite in nature. And this phenomenon of inducing a charge is known as charging by friction. The net charge on both of the bodies is 0 and the reason behind it is that an equal amount of charge repels it. When we rub a glass rod with a silk cloth, charge with opposite magnitude is generated over there. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.

Q 1.7)  (i) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?

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

Soln.:

(i) When a charge is placed in an electrostatic field then it experiences a continuous force. Therefore, an electrostatic field line is a continuous curve. And a charge moves continuously and does not jump from on point to the other. So, the field line cannot have a sudden break.

(ii) if two field lines will cross each other at any point then at that point the field intensity will start shooing two directions at the same point which is impossible. Therefore, two field lines can never cross each other.

Q 1.8) Two point charges qA = 3 µC and qB= –3 µC are located 20 cm apart in a vacuum.

(i) What is the electric field at the midpoint O of the line AB joining the two charges?

(ii) If a negative test charge of magnitude 1.5 × 10–9 C is placed at this point, what is the force experienced by the test charge?

Q 1.11) A polythene piece rubbed with wool is found to have a negative charge of 3 \times 10^{-7}\; C3×10−7C.

(i) Estimate the number of electrons transferred (from which to which?)

(ii) Is there a transfer of mass from wool to polythene?

Q 1.13) Suppose the spheres A and B in Exercise 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?

Question 1.15) Consider a uniform electric field E = 3 × 10 3 î N/C.

(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the y z – plane?

(b) What is the flux through the same square if the normal to its plane makes a 60 ° angle with the x-axis?

Soln. :

(a) Electric field intensity, E = 3 × 10 3 î  N / C

Magnitude of electric field intensity, | E | = 3 × 10 3 N / C

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

Area of the square, A = s 2 = 0.01 m 2

The plane of the square is parallel to the y – z plane. Hence, angle between the unit vector normal to the plane and electric field, θ = 0 °

Flux ( φ ) through the plane is given by the relation,

φ = | E |A cos θ

φ  = 3 × 10 3 × 0.01 × cos 0 °

φ = 30 N m 2 /C

(b) Plane makes an angle of 60 ° with the x – axis.

Hence, θ = 60°

Flux, φ = | E |A cos θ

Flux, φ  = 3 × 10 3 × 0.01 × cos 60°

Flux, φ  = 30 x 0.5

Flux, φ  = 15 N m 2 /C

Question 1.16)

What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

Soln.:

All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero.

Question 1.17)

Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 10 3 N m 2 /C.

(a) What is the net charge inside the box?

(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

Question 1.18)

A point charge + 10 μC is at a distance 5 cm directly above the centre of a square of side 10 cm, 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 10 cm )

Question 1.19)

A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Question 1.20)

A point charge causes an electric flux of – 1.0 × 10 3 N m 2 /C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge.

(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?

(b) What is the value of the point charge?

Question 1.21)

A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 10 3 N/C and points radially inward, what is the net charge on the sphere?

Question 1.22)

A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC /m 2

(a) Find the charge on the sphere.

(b) What is the total electric flux leaving the surface of the sphere?

Question 1.23) An infinite line charge produces a field of 9 × 10 4 N/C at a distance of 2 cm. Calculate the linear charge density.

Question 1.24) 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 × 10 – 22 C /m 2

What is E :

(a) In the outer region of the first plate,

(b) In the outer region of the second plate, and ( c ) between the plates?

Question 1.25) An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10NC–1 (Millikan’s oil drop experiment). The density of the oil is 1.26 g cm–3. Estimate the radius of the drop. (g = 9.81 m s–2; e = 1.60 × 10–19 C).

Question 1.27)

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 105 NC–1 per metre. 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?

Soln:

Total dipole moment of the system, p=q×dl=−10−7 cm

The rate of increase of the magnitude of the electric field along the positive z-direction = 105 NC–1 per metre.

The force experienced by the system is given as

F=qE

F=q(dE/dl) x dl
= p x (dE/dl)
=- 10−7 ×10−5

=−10−2N

The force experienced by the system is – 10−2 N. This is in the negative z-direction and opposite to the direction of the electric field. Therefore, the angle between the dipole moment and the electric field is 180o.
Torque (τ) =pEsin1800=0

Therefore, the torque experienced by the system is zero.

Question 1.28) (a) A conductor A with a cavity as shown in Fig. (a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor. (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Fig. (b)]. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Question 1.30)

Obtain the formula for the electric field due to a long thin wire of uniform linear charge density E without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

Soln:

Consider a long thin wire of uniform linear charge density λ. A point P is taken at a distance r from the midpoint C of the wire. The electric field at a point P due to the wire is E.

Let E be the electric field at point P due to the wire

Consider a small length element dx on the wire with centre O, such that OC = x

dx = rsec2θdθ

r2 + x2 = r2 + r2 tan2θ

Question 1.31) It is now established 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 + (2/3) e, and the ‘down’ quark (denoted by d) of charge (–1/3) 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.

Question 1.32)

(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show 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 a certain distance apart.

Soln:

(a) Let us consider that the equilibrium is stable. Then if the test charge is displaced in any direction it will experience a restoring force towards the null-point. This means that all the field lines near the null point will be directed towards the null point. That is, there is a net inward flux of electric field through a closed surface around the null-point. But according to Gauss’s law, the flux of electric field through a surface, not enclosing any charge, must be zero. Hence, equilibrium cannot be stable.
(b) The null-point is the mid-point of the line joining the two charges. If the test charge is displaced along the line from the null-point there is a restoring force. If the test charge is displaced normal to the line the net force takes it away from the null-point. Hence for the stability of equilibrium needs restoring force in all directions.

Question 1.33) A particle of mass m and charge (–q) enters the region between the two charged plates initially moving along x-axis with speed vx (like the figure in question 1.14). The length of the 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 qEL2/(2m vx2). Compare this motion with the motion of a projectile in a gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.

Question 1.34) Suppose that the particle in Exercise 1.33 is an electron projected with velocity v= 2.0 × 106 m s–1. If E between the plates separated by 0.5 cm is 9.1 × 102 N/C, where will the electron strike the upper plate? (|e|=1.6 × 10–19 C, me = 9.1 × 10–31 kg.)

NCERT Solutions for Class 12 Physics Chapter 1
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