Ncert Solution for Class 12 Physics Chapter 1 Electric Charges and Fields

Electric Charges and fields ncert solutions: Ncert Class 12 Physics Chapter 1 Exercise Solutions

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Question 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?

Solution 1.1:

Given:

• Charge on the first sphere,  q₁ = 2 × 10⁻⁷ C
• Charge on the second sphere, q₂ = 3 × 10⁻⁷ C
• Distance between the charges, r = 30 cm = 0.3 m

The force between the two charges is given by Coulomb’s law: \( F = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 q_2}{r^2} \)

Here, \( F = \frac{1}{4\pi \epsilon_0} \) is the Coulomb constant \( k = 9 \times 10^9 \, \text{Nm}^2\text{C}^{-2} \).

Substituting the values:

• \( F = \frac{9 \times 10^9 \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{(0.3)^2}\)
• \( F = \frac{9 \times 10^9 \times 6 \times 10^{-14}}{0.09} \)
• \( F = \frac{54 \times 10^{-5}}{0.09} = 6 \times 10^{-3} \, \text{N} \)

So, the force between the charges is \( 6 \times 10^{-3} \)N.

Hence, the force between the given charged particles will be \( 6 \times 10^{-3} \)N Since the nature of the charges is the same i.e. they are both positive. Hence, the force will be repulsive.

Question 1.2: The electrostatic force on a small sphere of charge is 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?

Solution 1.2:

Given:

• Charge on the first sphere, \( q_1 = 0.4 \, \mu C = 0.4 \times 10^{-6} \, C \)
• Charge on the second sphere, \( q_2 = -0.8 \, \mu C = -0.8 \times 10^{-6} \, C \)
• Force between the charges, \( F = 0.2 \, N \)

(a) The force between the two charges is given by Coulomb’s law: \( F = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 q_2}{r^2} \)

Rearranging for ( r ):

• \( r^2 = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 q_2}{F} \)
• \( r^2 = \frac{9 \times 10^9 \times 0.4 \times 10^{-6} \times 0.8 \times 10^{-6}}{0.2} \)
• \( r^2 = \frac{2.88 \times 10^{-2}}{0.2} = 1.44 \times 10^{-1} \, m^2 \)
• \( r = \sqrt{1.44 \times 10^{-1}} = 0.12 \, m = 12 \, cm \)

So, the distance between the two spheres is ( 0.12 )m or ( 12 )cm.

(b) The force on the second sphere due to the first will be the same in magnitude but opposite in direction due to Newton’s third law of motion. Hence, the force on the second sphere is also 0.2 N.

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

Solution 1.3:

Let’s check the dimensions:

• \( k = \frac{1}{4\pi \epsilon_0} has dimensions [M^1L^3T^{-2}A^{-2}]\)
• ( e ) (charge of an electron) has dimensions \([A^1T^1]\)
• ( G ) (gravitational constant) has dimensions \([M^{-1}L^3T^{-2}]\)
• ( m_e ) (mass of an electron) and ( m_p ) (mass of a proton) have dimensions \([M^1]\)

The expression for the ratio is: \( \frac{ke^2}{Gm_em_p}\)

Substituting the dimensions: \( \frac{[M^1L^3T^{-2}A^{-2}] \cdot [A^2T^2]}{[M^{-1}L^3T^{-2}] \cdot [M^1] \cdot [M^1]} = \frac{[M^1L^3T^{-2}A^{-2}] \cdot [A^2T^2]}{[M^2L^3T^{-2}]} = [1] \)

Hence, the ratio is dimensionless.

Now, calculate the value:

• \( k = 9 \times 10^9 \, \text{Nm}^2\text{C}^{-2} \)
• \( e = 1.6 \times 10^{-19} \, C \)
• \( G = 6.67 \times 10^{-11} \, \text{Nm}^2\text{kg}^{-2} \)
• \( m_e = 9.1 \times 10^{-31} \, kg \)
• \( m_p = 1.67 \times 10^{-27} \, kg \)

Substituting the values: \( \frac{ke^2}{Gm_em_p} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{6.67 \times 10^{-11} \times 9.1 \times 10^{-31} \times 1.67 \times 10^{-27}} \)

Simplifying this: \( = \frac{9 \times 10^9 \times 2.56 \times 10^{-38}}{1.013 \times 10^{-68}} = 2.3 \times 10^{39} \)

This ratio signifies the relative strength of the electrostatic force to the gravitational force between an electron and a proton.

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

Solution 1.4:

(a) Electric charge of a body is quantized. This means that only integral (1, 2, …., n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge.

(b) In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous.

Question 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.

Solution 1.5: Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed bodies is zero. This is because equal amounts of opposite charges annihilate each other. When a glass rod is rubbed with a silk cloth, opposite-natured charges appear on both bodies. This phenomenon is consistent with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.

Question 1.6: Four point charges \( q_A = 2 \, \mu\text{C} \), \( q_B = -5 \, \mu\text{C} ), (\ q_C = 2 \, \mu\text{C} \), and \( q_D = -5 \, \mu\text{C} \) are located at the corners of a square ( ABCD ) of side 10 cm. What is the force on a charge of \( 1 \, \mu\text{C} \) placed at the center of the square?

Solution 1.6: The given figure shows a square of a side 10 cm with four charges placed at its corners. O is the centre of the square.

(Sides) AB = BC = CD = AD = 10 cm

1. Calculate the distance from the center to each corner:

• The diagonal of the square is \( \sqrt{2} \times \text{side} = \sqrt{2} \times 10 \text{ cm} = 10\sqrt{2} \text{ cm} \).
• The distance from the center to a corner is \( \frac{\text{diagonal}}{2} = 5\sqrt{2} \text{ cm} = 0.05\sqrt{2} \text{ m} \).

2. Calculate the magnitude of the force exerted by each charge using Coulomb’s law: \( F = \frac{k \cdot |q \cdot Q|}{r^2} \)

where \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \), \( Q = 1 \, \mu\text{C} \), and \( r = 0.05 \sqrt{2} \text{ m} \).

The force magnitude for each charge is: \( F = \frac{8.99 \times 10^9 \cdot (2 \times 10^{-6})}{(0.05 \sqrt{2})^2} \)

Simplify to get: \(F = \frac{8.99 \times 10^9 \cdot 2 \times 10^{-6}}{0.0025 \times 2} = \frac{8.99\times 10^9 \times 2 \times 10^{-6}}{0.005}\)

\(F = \frac{17.98 \times 10^3}{0.005} \approx 3596 \text{ N}\)

3. Determine the net force considering the directions:

• Charges \( ( q_A ) and ( q_C ) are ( +2 \, \mu\text{C}) \) and attract.
• Charges \( ( q_B ) and ( q_D ) are ( -5 \, \mu\text{C}) \) and repel.

The forces from opposite charges cancel each other out due to symmetry, leading to: \(\text{Net force} = 0 \text{ N}\)

So, the net force on the \(1 \, \mu\text{C}\) charge at the center of the square is Zero N.

Question 1.7: (a) 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?

Solution 1.7: (a) An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.

(b) If two field lines cross each other at a point, then electric field intensity will show two directions at that point. This is not possible. Hence, two field lines never cross each other.

Question 1.8: Two point charges \( q_A = 3 \, \mu\text{C} \) and \( q_B = -3 \, \mu\text{C} \) are located \( 20 \, \text{cm} \) apart in vacuum. (a) What is the electric field at the midpoint \( O \) of the line \( AB \) joining the two charges? (b) If a negative test charge of magnitude \( 1.5 \times 10^{-9} \, \text{C} \) is placed at this point, what is the force experienced by the test charge?

Solution 1.8:

(a) The situation is represented in the given figure. O is the mid-point of line AB.

• Distance between two charges, AB = 20 cm
• Therefore, AO = OB = 10 cm
• The total electric field at the centre is (Point O) = E
• Electric field at point O caused by \( +3 \, \mu\text{C} \) charge,

\( E_{1} = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{3 \times 10^{-6}}{(OA)^{2}} = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{3 \times 10^{-6}}{(10 \times 10^{-2})^{2}} \, \text{N/C}\) along OB

Where ϵ0​ = Permittivity of free space

\( \frac{1}{4 \pi \epsilon_0} = 9 \times 10^{9} \; \text{Nm}^{2} \text{C}^{-2}\)

Therefore, The electric field at point O caused by \( -3 \, \mu\text{C} \) charge,

\(E_2 = \left| \frac{1}{4 \pi \epsilon_0} \cdot \frac{-3 \times 10^{-6}}{(OB)^2} \right| = \frac{1}{4 \pi \epsilon_0} \cdot \frac{3 \times 10^{-6}}{(10 \times 10^{-2})^2} \, \text{NC}\) along OB

\(\therefore E_{1} + E_{2} = 2 \times \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{3 \times 10^{-6}}{\left(10 \times 10^{-2}\right)^{2}} \; \text{NC}\) along OB

[Since the magnitudes of \( E_{1} \text{ and } E_{2}\) are equal and in the same direction]

\( \therefore E = 2 \times 9 \times 10^9 \times \frac{3 \times 10^{-6}}{\left(10 \times 10^{-2}\right)^2} \; \text{N/C}^{-1}\)

\( = 5.4 \times 10^6 \; \text{N/C}^{-1}\) along OB

Therefore, the electric field at midpoint O is \(5.4 \times 10^{6} \, \text{NC}^{-1}\) along OB

(b) A test charge with a charge potential of \( 1.5 \times 10^{-9} \, C\) is placed at midpoint O.

\(q = 1.5 \times 10^{-9} \; \text{C}\)

Let the force experienced by the test charge be F

Therefore, F = qE

• \(=1.5 \times 10^{-9} \times 5.4 \times 10^{6} \)
• \(= 8.1 \times 10^{-3} \; \text{N}\)

The force is directed along line OA because the negative test charge is attracted towards point A and is repelled by the charge placed at point B. As a result, the force experienced by the test charge is \(q = 8.1 \times 10 ^{-3} \; N\) along OA.

Question 1.9: A system has two charges \(q_A = 2.5 \times 10^{-7} \; \text{C} \quad \text{and} \quad q_B = -2.5 \times 10^{-7} \; \text{C}\) located at points A: (0, 0, -15 cm) and B: (0, 0, + 15 cm), respectively. What are the total charge and electric dipole moment of the system?

Solution 1.9: Both the charges can be located in a coordinate frame of reference as shown in the given figure.

• At point A, the total charge amount, \(q _{A} = 2.5 \times 10 ^{-7} \;C\)
• At point B, the total charge amount, \(q _{B} =  – 2.5 \times 10 ^{-7} \;C\)
• The total charge of the system is, \( q_A + q_B = 2.5 \times 10^{-7} \, \text{C} – 2.5 \times 10^{-7} \) C = 0
• Distance between two charges at points A and B, d = 15 + 15 = 30 cm = 0.3 m

The electric dipole moment of the system is given by,

• \(p = q_A \times d = q_B \times d \)
• \(= 2.5 \times 10^{-7} \times 0.3 \)
• \(= 7.5 \times 10^{-8} \; C\) m along the positive z−axis.

Therefore, the electric dipole moment of the system is \(7.5 \times 10^{-8} \; C\) m along positive z−axis.

Question 1.10: An electric dipole with dipole moment \(4 \times 10^{-9}\;  C\, m\) is aligned at 30° with the 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.

Solution 1.10:

• Electric dipole moment, \(p = 4 \times 10^{-9}\;C\,m\)
• Angle made by p with a uniform electric field, \(\theta = 30^{\circ}\)
• Electric field, \(E = 5 \times 10^{4}\; NC^{-1}\)
• The torque acting on the dipole is given by the relation, \(\tau = pE \sin \theta\ \)

\(= 4 \times 10^{-9} \times 5 \times 10^{4} \times \sin 30 \)

\(= 20 \times 10 ^{-5} \times \frac {1}{2} \)

\(= 10^{-4}\;Nm \)

Therefore, the magnitude of the torque acting on the dipole is \(10^{-4}\; Nm\).

Question 1.11: 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?) (b) Is there a transfer of mass from wool to polythene?

Solution 1.11: (a) When polythene is rubbed against wool, a number of electrons get transferred from wool to polythene. Hence, wool becomes positively charged and polythene becomes negatively charged.

• Charge on the polythene, q = \(3 \times 10^{-7}\; C\).
• Amount of charge on an electron,\(e = -1.6 \times 10^{-19}\; C\).
• Number of electrons transferred from wool to polythene = n

n can be calculated using the relation, q = ne

\(\Rightarrow n = \frac{q}{e} = \frac{-3 \times 10 ^{-7}}{-1.6 \times 10 ^{-19}} = 1.87 \times 10 ^{12}\).

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

(b) Yes. There is a transfer of mass taking place. This is because an electron has mass,

Mass of an electron, \(m_{e} = 9.1 \times 10 ^{-31} kg\).

Total mass transferred to polythene from wool,

• m = \(m_{e} \times n \ \).
• \(= 9.1 \times 10 ^{-31} \times 1.85 \times 10^{12} \ \).
• \(= 1.701 \times 10^{-18} \; kg\).

Hence, a negligible amount of mass is transferred from wool to polythene.

Question 1.12: 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 \(6.5 \times 10^{-7} \; C\) each? The radii of A and B are negligible compared to the distance of separation. (ii) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Solution 1.12: (a) Charge on sphere A, qA = Charge on sphere B, qB = \( 6.5 \times 10^{-7} \; C\)

Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between the two spheres, \(F = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_A q_B}{r^2}\)

Where, ϵ0 = Free space permittivity \(\frac{1}{ 4 \pi \epsilon _{0}} = 9 \times 10^{9} \; Nm^{2} C^{-2}\)

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

Therefore, the force between the two spheres is \(1.52 \times 10^{-2}\; N\)

(b) After doubling the charge, charge on sphere A, qA = Charge on sphere B, 2 × 6.5 × 10−7 C = 1.3 × 10−6 C

The distance between the spheres is halved. \(∴ r = \frac {0.5}{2} = 0.25 \; m\)

Force of repulsion between the two spheres,

\( F = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_A q_B}{r^2} = \frac{9 \times 10^9 \times 1.3 \times 10^{-6} \times 1.3 \times 10^{-6}}{(0.25)^2} \)

\( = 16 \times 1.52 \times 10^{-2} \)

\( = 0.243 \; N \)

Therefore, the force between the two spheres is 0.243 N.

Question 1.13: The 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?

Solution 1.13: Opposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged.

The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.

Question 1.14: Consider a uniform electric field E = 3 × 10³ 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?

Solution 1.14: (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, the 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 1.15: 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?

Solution 1.15: 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 1.16: 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 Nm² /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?

Solution 1.16: (a) Net outward flux through the surface of the box, φ = 8.0 × 10³ N m² /C

For a body containing net charge q, flux is given by the relation,

\( \phi = \frac{ q }{ \varepsilon _{ 0 }}\)

\( \varepsilon _{ 0 } = permitivity of free space\)

\( \varepsilon _{ 0 } = 8.854 \times 10 ^{ – 12 } N^{ – 1 } C ^{ 2 } m ^{ – 2 }\)

\( q = \varepsilon _{ 0 } \phi\)

= 8.854 x 10^–12 x 8.0 x 10³ C = 7.08 x 10^–8 C = 0.07 \(\mu C\)

Thus, the total charge inside the box is 0.07 \( \mu C\)

(b) No

Net flux piercing out through a body depends on the net charge contained in the body. If the net flux is zero, then it can be inferred that the net charge inside the body is zero. The body may have an equal amount of positive and negative charges.

Question 1.17: A point charge + 10 \( \mu C\) 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.)

Solution 1.17: The square can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed. According to Gauss’s theorem for a cube, total electric flux is through all its six faces.

\(\phi_{\text{Total}} = \frac{q}{\epsilon_0}\)

Hence, electric flux through one face of the cube, i.e. through the square, is \(\phi = \frac{\phi_{\text{Total}}}{6} = \frac{1}{6} \cdot \frac{q}{\epsilon_{0}}\)

Here, \({ \epsilon _{ 0 }}\)

= permittivity of free space

= 8.854 x 10^–12 N^–1 C² m^–2

\(q = 10 \mu C = 10 \times 10 ^{ – 6 } C\)

Therefore, \(\phi = \frac{ 1 }{ 6 } . \frac{ 10 \times 10 ^{ – 6 }}{ 8.854 \times 10 ^{ – 12 }}\)

= 1.88 x 10⁵ N m² C^–1

Therefore, electric flux through the square is 1.88 × 10⁵ N m² C^–1

Question 1.18: A point charge of 2.0 \( \mu C\) is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Solution 1.18: Net electric flux ( φ net) through the cubic surface is given by \(\phi_{\text{net}} = \frac{q}{\epsilon_0}\)

Here, \({ \epsilon _{ 0 }}\)

= permittivity of free space

= 8.854 × 10^–12 N^–1 C² m^–2

\(q = net \;charge\; contained \;in \;the\; cube\; given \;\)

\(= \;2.0 \mu\; C \;=\; 2 \times 10 ^{ – 6 } C\)

Therefore, \(\phi _{ net } = \frac{ 2 \times 10 ^{ – 6 }}{ 8.854 \times 10 ^{ – 12 }}\)

\(\phi _{ net } = 2.26 \times 10 ^{ 5 } N m ^{ 2 } C ^{ – 1}\)

The total electric flux through the surface of the cube given is, \(\phi _{ net } = 2.26 \times 10 ^{ 5 } N m ^{ 2 } C ^{ – 1}\)

Question 1.19: A point charge causes an electric flux of –1.0 × 10³ Nm² /C  to pass through a spherical Gaussian surface of a 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?

Solution 1.19: (a) Electric flux, Φ = – 1.0 × 10³ N m² /C

The radius of the Gaussian surface, r = 10.0 cm

Electric flux penetrating through a surface depends on the net charge enclosed inside the body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same, i.e., – 10³ N m² /C.

(b) Electric flux is given by the relation

\(\phi_{\text{Total}} = \frac{q}{\epsilon_0}\)

Here, \({ \epsilon _{ 0 }}\)

= permittivity of free space = 8.854 x 10^–12 N^–1 C ² m^–2

q = total charge contained enclosed by the spherical surface = φ ε ₀

q = – 1.0 x 10³ x 8.854 x 10^–12

q = – 8.854 x 10^–9 C

q = – 8.854 nC

Therefore, the value of the point charge is – 8.854 nC.

Question 1.20: 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 N/C and points radially inward, what is the net charge on the sphere?

Solution 1.20: Electric field intensity (E) at a distance (d) from the centre of a sphere containing net charge q is given by the relation,

\(E = \frac{ q }{ 4\pi \epsilon _{ 0 } d ^{ 2 }}\)

Where, q = Net charge = 1.5 × 10³ N/C

d = Distance from the centre = 20 cm = 0.2 m

ε₀ = Permittivity of free space and

\(\frac{ 1 }{ 4 \pi \epsilon _{ 0 }} = 9 \times 10 ^{ 9 } N m ^{ 2 } C ^{ – 2 }\)

Therefore, the net charge on the sphere is 6.67 n C.

Question 1.21: A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 \( \mu C\)/m 2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?

Solution 1.21: (a) The diameter of the sphere, d = 2.4 m

The radius of the sphere, r = 1.2 m

Surface charge density, σ = 80.0 \( \mu C\)/m² = 80 × 10^–6 C/m²

The total charge on the surface of the sphere can be calculated as follows,

Q = Charge density × Surface area

= σ × 4πr ²

= 80 × 10^–6 × 4 × 3.14 × ( 1.2 )²

= 1.447 × 10^–3 C

Therefore, the charge on the sphere is 1.447 × 10^–3 C.

(b) Total electric flux ( φ _Total  ) leaving out the surface of a sphere containing net charge Q is given by the relation,

\(\phi_{\text{Total}} = \frac{q}{\epsilon_0}\)

Here, \({ \epsilon _{ 0 }}\)

= permittivity of free space  = 8.854 x 10^–12 N^–1 C² m^–2

Electric flux = 1.63 x 10⁸ N C^–1 m²

Therefore, the total electric flux leaving the surface of the sphere is 1.63 × 10⁸ N C^–1 m²

Question 1.22: An infinite line charge produces a field of 9 × 10⁴ N/C at a distance of 2 cm. Calculate the linear charge density.

Solution 1.22: The electric field produced by the infinite line charges at a distance d having linear charge density λ is given by the relation,

\(E = \frac{ \lambda }{ 2 \pi \epsilon _{ 0 } d }\)

Here,

d = 2 cm = 0.02 m

E = 9 x 10⁴ N / C

ε ₀ = Permittivity of free space and

\(\frac{ 1 }{ 4 \pi \epsilon _{ 0 }} = 9 \times 10 ^{ 9 } N m ^{ 2 } C ^{ – 2 }\)

\(\ \lambda = \frac{0.02\times 9\times 10^{4} }{2\times 9\times 10^{9} } \ \lambda = 10 \mu C/m\)

The linear charge density is \(10 \mu C/m\).

Question 1.23: 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 magnitude 17.0 × 10^–22 C /m² 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?

Solution 1.23: The situation given in the question can be depicted in the following figure.

Parallel plates A and B are placed close to each other, as shown in the figure. The region between plates A and B is labelled as II, the outer region of plate A is labelled as I, and the outer region of plate B is labelled as III.

(a) The charge density of plate A can be calculated as,

σ = 17.0 × 10^–22 C/m²

(b) Similarly, the charge density of plate B can be calculated as,

σ = – 17.0 × 10^–22 C/m²

(c) In regions I and III, electric field E is zero. This is because the charge is not enclosed by the respective plates.

Electric field E in region II is given by the relation,

\(E = \frac{\sigma }{ \epsilon _{ 0 }}\)

ε ₀ = Permittivity of free space = 8.854 × 10^–12 N^–1C² m^–2

\(E = \frac{ 17.0 \times 10 ^{ – 22 }}{ 8.854 \times 10 ^{ – 12 }}\)

= 1.92 x 10^–10 N/C

Therefore, the electric field between the plates is 1.92 x 10^–10 N/C