NCERT EXERCISES WITH SOLUTION CHAPTER-1 ELECTRIC CHARGES AND FIELDS

 

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

Ans. Here q₁ = 2 × 10^-7C, 

q₂ = 3 × 10^-7C,

r = 30 cm = 0.30 m

According to Coulomb’s law,

  (Repulsive).

1.2. The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge -0.8μC in air is 0.2 N. (i) What is the distance between two spheres ? (ii) What is the force on the second sphere due to the first ?

Ans. (i) Here q₁ = 0.4 μC = 0.4 × 10^-6 C

q₂ = – 0.8 μC = – 0.8 × 10^-6 C, F = 0.2 N, r = ?

As 

∴  

or  cm.

(ii) The two charges mutually exert equal and opposite forces.

∴ Force on the second sphere due to the first = 0.2 N (attractive).

1.3. Check that the ratio ke²/Gm_em_p is dimensionless. Look up a table of physical constants and determine the value of this ratio. What does this ratio signify?

Ans.  unit

As the ratio k e² / Gmm has no unit, so it is dimensionless.

Now k = 9 × 10⁹Nm²C^-2

G = 6.67 × 10^-11 Nm² kg^-2

 e = 1.6 × 10^-19 kg

m_e = 9.1 × 10^-31 kg

and m_p = 1.66 × 10^-27 kg

∴  

= 2.287 ×10³⁹.

The factor ke² / Gm_em_p represents the ratio of electrostatic force to the gravitational force between an electron and a proton. Also, the large value of the ratio signifies that the electrostatic force is much stronger than the gravitational force.

1.4. (i) Explain the meaning of the statement ‘electric charge of a body is quantized.’

(ii) Why can one ignore quantization of electric charge when dealing with macroscopic i.e., large scale charges ?

Ans. (i) Quantization of electric charge means that the total charge (q) of a body is always an integral multiple of a basic charge (e) which is the charge on an electron. Thus q = ne, where n = 0, ± 1, ± 2, ± 3, ……              

(ii) While dealing with macroscopic charges (q = ne), we can ignore quantization of electric charge. This is because e is very small and n is very large and so q behaves as if it were continuous i.e., as if a large amount of charge is flowing continuously.

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.

Ans. It is observed that the positive charge developed on the glass rod has the same magnitude as the negative charge developed on silk cloth. So total charge after rubbing is zero as before rubbing. Hence the law of conservation of charge is being obeyed here.

1.6. Four point charges q_A = 2 μC, q_B = -5 μC, q_C = 2 μC, q_D = -5 μC are located at the corners of a square ABCD of side

Fig. 1.149

10 cm. What is the force on a charge of 1 μC placed at the centre of the square ?

Ans. Here 

 cm 

Forces exerted on the charge of 1 μC located at the centre are

 , along 

 , along 

 , along 

 , along 

Clearly,  and 

Hence total force on 1 μC charge is

  zero N.

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

(b) Explain why two field lines never cross each other at any point ? [Punjab 01, 02; CBSE D 05, 03 ; OD 14]

Ans. (a) Electric lines of force exist throughout the region of an electric field. The electric field of a charge decreases gradually with increasing distance from it and becomes zero at infinity i.e., electric field cannot vanish abruptly. So a line of force cannot have sudden breaks, it must be a continuous curve.

(b) If two lines of force intersect, then there would be two tangents and hence two directions of electric field at the point of intersection, which is not possible.

1.8. Two point charges q_A = + 3 μC and q_B = -3 μC are located 20 cm apart in vacuum, (i) Find 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 the centre, find the force experienced by the test charge.                                           [CBSE OD 03]

Ans. The directions of the fields E_A and E_B due to the charges q_A and q_B at the midpoint P are as shown in Fig. 1.150.

Electric field at the midpoint O due to q_A,

 , along 

Electric field at the midpoint O due to q_B,

 , along 

Resultant field at the midpoint  is

 , along 

(ii) Force on a negative charge of 1.5 × 10^-9C placed at the midpoint O,

 , along 

The force on a negative charge acts in a direction opposite to that of the electric field.

1.9. A system has two charges q_A = 2.5 ×10^-7 C and q_B = -2.5 × 10^-7 C, located at points A (0,0, -15 cm) and B (0,0, +15 cm) respectively. What is the total charge and electric dipole moment of the system ?

Ans. Clearly, the two charges lie on Z-axis on either side of the origin and at 15 cm from it, as shown in Fig. 

∴ 2a = 30 cm = 0.30 m, q = 2.5 × 10^-7 C

Total charge = q_A + q_B = 2.5 × 10^-7 – 2.5 × 10^-7 = 0

Dipole moment,

p – q × 2a = 2.5 × 10^-7 × 0.30

 = 0.75 × 10^-7 Cm

The dipole moment acts in the direction from B to A i.e., along negative Z-axis.

1.10. An electric dipole with dipole moment 4 × 10^-9 Cm is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 10⁴ NC^-1 Calculate the magnitude of the torque acting on the dipole.

Ans. Here p = 4 × 10“⁹ Cm, θ = 30°, E = 5 × 10⁴ NC^-1

∴ Torque,  = pE sin θ

= 4 × 10^-9 × 5 ×10⁴ × sin 30°

= 10^-4 Nm.

1.11. A polythene piece rubbed with wool is found to have a negative charge of 3.2 × 10^-7C. (i) Estimate the number of electrons transferred, (ii) Is there a transfer of mass from wool to polythene ?

Ans. (i) Here q = 3.2 × 10^-7 C, e = 1.6 × 10^-19 C

As q = ne, therefore

Number of electrons transferred,

 

Since polythene has negative charge, so electrons are transferred from wool to polythene during rubbing.

(ii) Yes, there is a transfer of mass from wool to polythene because each electron has a finite mass of 9.1 × 10^-31 kg.

Mass transferred

= m_e × n = 9.1 × 10^-31 × 2 × 10¹²

= 1.82 × 10^-18 kg

Clearly, the amount of mass transferred is negligibly small.

1.12. (a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10^-7C ? The radii of A and B are negligible compared to the distance of separation, (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved ?

Ans. 

1.13. 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 ?

Ans. 

1.14. Figure 1.152 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

1.15. Consider a uniform electric field : = 3 × 10³ NC^-1 (i) 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 ? (ii) What is the flux through the same square if the normal to its plane makes a 60° angle with the X-axis?

Ans. (i) Normal to a plane parallel to Y-Z plane points in X-direction, so

 = 0.10 × 0.10  m² = 0.01  m²

Electric flux,

ϕ_E =  .  = 3 × 10³  . 0.01 

= 30 .  = 30 Nm²C^-1.

(ii) Here θ = 60°

∴ ϕ_E = E∆S cos 60° = 3 × 10³ × 0.01 cos 60°

= 30 ×  = 15 Nm²C^-1.

1.16. Consider a uniform electric field : = 3 × 10³ NC^-1. What is the net flux of this field through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

Ans. The flux entering one face parallel to Y-Z plane is equal to the flux leaving other face parallel to Y-Z plane. Flux through other faces is zero. Hence net flux through the cube is zero.

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 × 10³ Nm²C¹. (i) What is the net charge inside the box ? (ii) 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 ?

Ans. (i) ϕ_E = 8.0 × 10³ Nm²C^-2

Using Gauss theorem,

Charge, q = ε₀. ϕ_E = 8.0 × 10³ × 

= 0.07 × 10^-6 = 0.07 μC

(ii) No, we cannot say that there are no charges at all inside the box. We can only say that the net charge inside the box is zero.

1.18. A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm as shown in Fig. 1.153(a). 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)

Ans. We can imagine the square as face of a cube with edge 10 cm and with the charge of + 10 μC placed at its centre, as shown in 

Symmetry of six faces of a cube about its centre ensures that the flux ϕ_S through each square face is same when the charge q is placed at the centre.

∴ Total flux,

ϕ_E = 6 × ϕ_S = 

or ϕ_S =  × 10 × 10^-6 × 4π × 9 × 10⁹

= 1.88 ×10⁵ Nm²C^_1.

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 ?

Ans. Here q = 2.0 μ C = 2.0 × 10^-6C,

ε₀ = 8.85 × 10^-12C²N^-1m^-2

By Gauss’s theorem, electric flux is

 = 2.26 × 10⁵ Nm² C^-1.

1.20. A point charge causes an electric flux of -1.0 × 10³ Nm² C^-1 to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge, (i) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface ? (ii) What is the value of the point charge ?

Ans. (i) ϕ_E = -10³ Nm²C^-1, because the charge enclosed is the same in both the cases.

(ii) Charge,

q = ε₀ ϕ_E

=  ×(- 1.0 × 10³)

= – 8.84 × 10^-9 C = – 8.84 nC.

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 × 10³ NC^-1 and points radially inward, what is the net charge on the sphere ?

Ans. Electric field at the outside points of a conducting sphere is

E = 

∴ q = 4πε₀Er² =  × 1.5 × 10³ × (0.20)² C

= 6.67 × 10^-9 C = 6.67 nC

As the field acts inwards, the charge q must be negative.

∴  q = – 6.67 nC.

1.22. A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m². (i) Find the charge on the sphere, (ii) What is the total electric flux leaving the surface of the sphere ? [CBSE D 09C]

Ans. Here R =  1.2 m

σ = 80.0μCm^-2 = 80 × 10-6 Cm^-2

(i) Charge on the sphere is

q = 4π R² σ = 4 × 3.14 ×(1.2)² × 80 × 10^-6C

= 1.45 ×10^-3 C.

(ii) Flux,

ϕ_E =  = 1.45 × 10^-3 × 4π × 9 × 10⁹

= 1.6 ×10⁸ Nm² C^-1.

1.23. An infinite line charge produces a field of 9 × 10⁴ NC^-1 at a distance of 2 cm. Calculate the linear charge density.

Ans. E = 9 × 10⁴ NC^-1, r = 2cm = 0.02 m

Electric field of a line charge, E = 

∴ Linear charge density,

= 2πε₀Er = 2π × × 9 × 10⁴ × 0.02

= 0.01 × 10^-5 Cm^-1 = 0.1 μCm^-1.

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 Cm^-2. What is E (a) to the left of the plates, (b) to the right of the plates, and (c) between the plates ?

Ans. Here σ = 17.0 × 10^-22 Cm^-2

(a) On the left, the fields of the two plates are equal and opposite, so E = Zero.

(b) On the right, the fields of the two plates are equal and opposite, so E = Zero.

(c) Between the plates, the fields due to both plates are in same direction. So the resultant field is

 

= 19.2 × 10^-10 NC^-1.

1.25. An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10⁴ Vm^-1 in 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 ms^-2; e = 1.60 × 10^-19 C)

Ans. Force on the oil drop due to electric field = qE = neE

Weight of oil drop

= mg = volume × density × g =  π r³g

The field E must act vertically downward so that the negatively charged oil drop experiences an upward force and balances the weight of the drop.

When the drop is held stationary,

Weight of oil drop

= Force on the oil drop due to electric field

or    ∴ 

Now,  

 

  kg 

∴  

 mm.

1.26. Which among the curves shown in Fig. 1.155, cannot possibly represent electrostatic field lines ?

Ans. Only Fig. (c) is right and the remaining figures cannot represent the electrostatic field lines.

Figure (a) is wrong because field lines must be normal to a conductor.

Figure (b) is wrong because lines of force cannot start from a negative charge.

 

 

 

 

Figure (c) is right because it satisfies all the properties of lines of force.

Figure (d) is wrong because lines of force cannot intersect each other.

Figure (c) is wrong because electrostatic field lines cannot form closed loops.

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^-1m^-1. What are the force and torque experienced by a system having a total dipole moment equal to 10^-7 C m in the negative Z-direction ?

Ans. As the electric field changes uniformly in the positive Z-direction, so

As the system has a total dipole moment in the negative Z-direction, so

P_z = – 10^-7 Cm, p_x = 0, p_y = 0

In a non-uniform electric field, the force on the dipole will be

= 0 + 0 – 10^-7 ×10⁵ = -10^-2 N

The negative sign shows that the force on the dipole acts in the negative Z-direction.

As the dipole moment p acts in the negative Z-direction while the electric field E acts in the positive Z-direction, so θ = 180°.

Torque,  = pE sin 180° = pE × 0 = 0.

1.28. (i) A conductor A with a cavity [Fig. (a)] is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.

(ii) 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)],

(iii) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

 

Ans. (i).

(ii) Consider a Gaussian surface inside the conductor but quite close to the cavity.

Inside the conductor, E = 0.

By Gauss’s theorem,

i.e., the total charge enclosed by the Gaussian surface must be zero. This requires a charge of -q units to be induced on inner surface of conductor A. But an equal and opposite charge of +q units must appear on outer surface A so that charge on the surface of A is Q + q.

Hence the total charge on the surface of A is Q + q.

(iii) The instrument should be enclosed in a metallic case. This will provide an electrostatic shielding to the instrument.

1.29. A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is  , where  is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.

Ans. Consider the charged conductor with the hole filled up, as shown by shaded portion in Fig. 1.159. Applying Gauss’s theorem, we find that field just outside is   and is zero inside. This field can be viewed as the superposition of the field E₂ due to the filled up hole plus

the field due to the rest of the charged conductor. Since inside the conductor the field vanishes, the two fields must be equal and opposite, i.e.,

E₁ – E₂ = 0    …(1)

And outside the conductor, the fields are added up :

   …(2)

Adding equations (1) and (2), we get

  or  

Hence the field due to the rest of the conductor or the field in the hole is

where n is a unit vector in the outward normal direction.

1.30. Obtain the formula for the electric field due to a long thin wire of uniform linear charge density  without using Gauss’s law.

[Hint. Use Coulomb’s law directly and evaluate the necessary integral.]

Ans. 

1.31. It is now believed that protons and neutrons are themselves built out of more elementary units called quarks. A proton and a neutron consists 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. Suggest a possible quark composition of a proton and neutron.

Ans. Charge on ‘up^’ quark (u) = +  e

Charge on ‘down^’ quark (d) =  e

Charge on a proton = e

Charge on a neutron = 0

Let a proton contain x ‘up’ quarks and (3 – x) ‘down’ quarks. Then total charge on a proton is

ux + d (3 – x) = e

or 

or 

or    x = 2 and 3 – x = 3 – 2 = 1

Thus a proton contains 2 ‘up’ quarks and 1 ‘down’ quark. Its quark composition should be Let a neutron contain y ‘up’ quarks and (3 – y) ‘down’ quarks. Then total charge on a neutron must be

uy + d (3 – y) = 0

or 

or  

or y = 1 and 3 – y = 3 – 1 = 2

Thus a neutron contains 1 ‘up’ quark and 2 ‘down’ quarks. Its composition should be : 

1.32. (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where  = 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.

Ans. (a) We can prove it by contradiction. Suppose the test charge placed at null point be in stable equilibrium. Since the stable equilibrium requires restoring force in all directions, therefore, the test charge displaced slightly in any direction will experience a restoring force towards the null point. That is, all field lines near the null point should be directed towards the null point. This indicates that there is a net inward flux of electric field through a closed surface around the null point. But, by Gauss’s law, the flux of electric field through a surface enclosing no charge must be zero. This contradicts our assumption. Hence the test charge placed at the centre must be necessarily in unstable equilibrium.

(b) The null point lies on the midpoint of the line joining the two charges. If the test charge is displaced slightly on either side of the null point along this line, it will experience a restoring force. But if it is displaced normal to this line, the net force takes it away from the null point. That is, no restoring force acts in the normal direction. But stable equilibrium demands restoring force in all directions, hence test charge placed at null point will not be in stable equilibrium.

1.33. A particle of mass mand charge (- q) enters the region between the two charged plates initially moving along x-axis with speed v_x (like particle 1 in Fig. 1.152). The length of plate is Land 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 ).

Compare this motion with motion of a projectile in gravitational field.

Ans. The motion of the charge – q in the region of the electric field E between the two charged plates is shown in 

Force on the charge – q in the upward direction is

 ma = qE

∴ Acceleration, 

Time taken to cross the field, t = 

Vertical deflection at the far edge of the plate will be

Like the motion of a projectile in gravitational field, the path of a charged particle in an electric field is parabolic.

1.34. Suppose that the particle in Exercise 1.33 is an electron projected with velocity v_x = 2.0 × 10⁶ ms^-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, m_e = 9.1 × 10^-31 kg).

Ans. Here y = 0.5 cm = 0.5 × 10^-2 m,

v_× = 2.0 × 10⁶ ms^-1, E = 9.1 × 10² NC^-1, L = ?

From the above exercise, the vertical deflection of an electron is given by

∴ 

or   cm.