Electrostatic Potential and Capacitance Chapter-Wise Test 9

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A parallel plate capacitor with capacitance \( 120 \, \text{pF} \) has a dielectric (\( K = 4 \), thickness \( d/4 \)) inserted. What is the new capacitance? (Original separation \( d \)).

Potential difference: \( V = E_0 \left( \frac{3d}{4} \right) + \frac{E_0}{K} \left( \frac{d}{4} \right) = E_0 d \left( \frac{3}{4} + \frac{1}{4 \times 4} \right) \).

\( V = E_0 d \left( \frac{3}{4} + \frac{1}{16} \right) = E_0 d \times \frac{13}{16} \).

\( C = \frac{Q}{V} = \frac{Q}{\frac{13}{16} V_0} = \frac{16}{13} \times 120 \approx 147.69 \, \text{pF} \).

140 pF
145 pF
150 pF
147.69 pF
4

A point charge \( Q = 24 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 8 m away from the charge. (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Potential due to a point charge: \( V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r} \).

Substitute: \( V = 9 \times 10^9 \times \frac{24 \times 10^{-9}}{8} = 9 \times 10^9 \times 3 \times 10^{-9} = 27 \, \text{V} \).

27 V
30 V
24 V
20 V
1

A spherical conductor of radius 25 cm has a charge of \( 10 \times 10^{-8} \, \text{C} \). What is the electric field at 60 cm from the center? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

For \( r = 0.6 \, \text{m} > R = 0.25 \, \text{m} \), \( E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2} \).

\( E = 9 \times 10^9 \times \frac{10 \times 10^{-8}}{(0.6)^2} = 9 \times 10^9 \times \frac{10 \times 10^{-8}}{0.36} = 2.5 \times 10^3 \, \text{N/C} \).

2 × 10³ N/C
3 × 10³ N/C
2.5 × 10³ N/C
4 × 10³ N/C
3

A capacitor of \( 5 \, \mu\text{F} \) is charged to \( 100 \, \text{V} \). What is the energy stored in it?

\( U = \frac{1}{2} C V^2 = \frac{1}{2} \times 5 \times 10^{-6} \times (100)^2 = \frac{1}{2} \times 5 \times 10^{-6} \times 10^4 = 0.025 \, \text{J} \).

0.02 J
0.025 J
0.03 J
0.05 J
2

An electric dipole with moment \( p = 6 \times 10^{-10} \, \text{C m} \) lies along the x-axis. What is the potential at \( (5, 0, 0) \, \text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Along the dipole axis (\( \theta = 0^\circ \)): \( V = \frac{1}{4 \pi \varepsilon_0} \frac{p}{r^2} \).

\( V = 9 \times 10^9 \times \frac{6 \times 10^{-10}}{5^2} = 9 \times 10^9 \times \frac{6 \times 10^{-10}}{25} = 2.16 \, \text{V} \).

2.16 V
1.8 V
2.5 V
3 V
1

A dipole with \( p = 6 \times 10^{-9} \, \text{C m} \) makes an angle of \( 45^\circ \) with a uniform field \( E = 5 \times 10^4 \, \text{N/C} \). What is its potential energy?

\( U = -p E \cos \theta = -6 \times 10^{-9} \times 5 \times 10^4 \times \cos 45^\circ \).

\( \cos 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707 \), so \( U = -6 \times 10^{-9} \times 5 \times 10^4 \times 0.707 = -2.121 \times 10^{-4} \, \text{J} \).

-2.5 × 10⁻⁴ J
-2.121 × 10⁻⁴ J
-2 × 10⁻⁴ J
-1.5 × 10⁻⁴ J
2

Why does the potential energy of an electric dipole in a uniform electric field become minimum when the dipole aligns with the field?

The potential energy of a dipole in a uniform field is \( U = -p E \cos \theta \), where \( p \) is the dipole moment, \( E \) is the field strength, and \( \theta \) is the angle between \( p \) and \( E \). The energy is minimum when \( \cos \theta = 1 \), i.e., \( \theta = 0^\circ \), meaning the dipole aligns with the field. At this position, the torque (\( \tau = p E \sin \theta \)) is zero, and the system is in stable equilibrium, minimizing the potential energy.

The field exerts maximum torque
The dipole moment cancels the field
The field becomes non-uniform
The torque is zero at alignment
2

A parallel plate capacitor with capacitance \( 60 \, \text{pF} \) has a dielectric (\( K = 4 \), thickness \( d/4 \)) inserted. What is the new capacitance? (Original separation \( d \)).

Potential difference: \( V = E_0 \left( \frac{3d}{4} \right) + \frac{E_0}{K} \left( \frac{d}{4} \right) = E_0 d \left( \frac{3}{4} + \frac{1}{4 \times 4} \right) = E_0 d \left( \frac{3}{4} + \frac{1}{16} \right) = E_0 d \times \frac{13}{16} \).

\( C = \frac{Q}{V} = \frac{Q}{\frac{13}{16} V_0} = \frac{16}{13} \times 60 \approx 73.85 \, \text{pF} \).

70 pF
72 pF
75 pF
73.85 pF
4

Two capacitors of \( 60 \, \text{pF} \) and \( 120 \, \text{pF} \) are connected in series. What is the equivalent capacitance?

\( \frac{1}{C} = \frac{1}{60} + \frac{1}{120} = \frac{2 + 1}{120} = \frac{3}{120} \).

\( C = \frac{120}{3} = 40 \, \text{pF} \).

35 pF
40 pF
45 pF
50 pF
2

A dipole with \( p = 3 \times 10^{-10} \, \text{C m} \) makes an angle of \( 90^\circ \) with a uniform field \( E = 2 \times 10^5 \, \text{N/C} \). What is its potential energy?

\( U = -p E \cos \theta = -3 \times 10^{-10} \times 2 \times 10^5 \times \cos 90^\circ = 0 \, \text{J} \).

-6 × 10⁻⁵ J
0 J
6 × 10⁻⁵ J
3 × 10⁻⁵ J
2

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