Electrostatic Potential and Capacitance Chapter-Wise Test 14

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Two charges \( 20 \, \mu\text{C} \) and \( -10 \, \mu\text{C} \) are at \( (1, 0, 0) \) and \( (-1, 0, 0) \, \text{cm} \). What is the potential at midpoint? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Distance to midpoint = 0.01 m.

\( V = 9 \times 10^9 \left( \frac{20 \times 10^{-6}}{0.01} + \frac{-10 \times 10^{-6}}{0.01} \right) = 9 \times 10^9 \times \frac{10 \times 10^{-6}}{0.01} \).

\( V = 9 \times 10^9 \times \frac{10 \times 10^{-6}}{0.01} = 9 \times 10^6 \, \text{V} \).

8.5 × 10⁶ V
8.8 × 10⁶ V
9.2 × 10⁶ V
9 × 10⁶ V
4

A charged particle is placed in a region where the electric potential is constant everywhere. What can be inferred about the motion of the particle if it is released from rest?

If the electric potential \( V \) is constant everywhere, the electric field \( E = -\frac{dV}{dx} = 0 \), since the gradient of a constant potential is zero. With no electric field, there is no electrostatic force (\( F = qE \)) acting on the charged particle. If released from rest, the particle experiences no acceleration due to the electric field and thus remains at rest, assuming no other forces (like gravity) are considered.

It remains at rest
It moves in a circular path
It accelerates uniformly
It oscillates periodically
1

A parallel plate capacitor with capacitance \( 50 \, \text{pF} \) has a dielectric (\( K = 2 \), thickness \( d/3 \)) inserted. What is the new capacitance? (Original separation \( d \)).

Potential difference: \( V = E_0 \left( \frac{2d}{3} \right) + \frac{E_0}{K} \left( \frac{d}{3} \right) = E_0 d \left( \frac{2}{3} + \frac{1}{3 \times 2} \right) = E_0 d \left( \frac{2}{3} + \frac{1}{6} \right) = E_0 d \times \frac{5}{6} \).

\( C = \frac{Q}{V} = \frac{Q}{\frac{5}{6} V_0} = \frac{6}{5} \times 50 = 60 \, \text{pF} \).

55 pF
58 pF
62 pF
60 pF
4

A dipole \( p = 4 \times 10^{-10} \, \text{C m} \) is in a uniform field \( E = 10^5 \, \text{N/C} \) at \( 60^\circ \). What is its potential energy?

\( U = -\mathbf{p} \cdot \mathbf{E} = -p E \cos \theta = -4 \times 10^{-10} \times 10^5 \times \cos 60^\circ = -4 \times 10^{-5} \times 0.5 = -2 \times 10^{-5} \, \text{J} \).

-1 × 10⁻⁵ J
-2 × 10⁻⁵ J
-3 × 10⁻⁵ J
-4 × 10⁻⁵ J
2

A dipole with \( p = 3 \times 10^{-9} \, \text{C m} \) makes an angle of \( 60^\circ \) with a uniform field \( E = 2 \times 10^5 \, \text{N/C} \). What is its potential energy?

\( U = -p E \cos \theta = -3 \times 10^{-9} \times 2 \times 10^5 \times \cos 60^\circ \).

\( \cos 60^\circ = 0.5 \), so \( U = -3 \times 10^{-9} \times 2 \times 10^5 \times 0.5 = -3 \times 10^{-4} \, \text{J} \).

-4 × 10⁻⁴ J
-3 × 10⁻⁴ J
-2 × 10⁻⁴ J
-1 × 10⁻⁴ J
2

Why does the electric field inside a conductor vanish in electrostatic equilibrium, even if the conductor is irregularly shaped?

In electrostatic equilibrium, free charges in a conductor redistribute to cancel any internal electric field. If there were a field inside, it would exert a force on the free charges, causing them to move until the field becomes zero everywhere inside. This principle holds regardless of shape because conductors allow charge mobility, and equilibrium requires \( E = 0 \) inside, with all excess charge residing on the surface, adjusting to the geometry.

Charges redistribute to cancel the internal field
The shape prevents field penetration
The conductor absorbs the external field
The field accumulates on the surface only
1

A parallel plate capacitor with capacitance \( 150 \, \text{pF} \) has a dielectric (\( K = 6 \), thickness \( d/5 \)) inserted. What is the new capacitance? (Original separation \( d \)).

Potential difference: \( V = E_0 \left( \frac{4d}{5} \right) + \frac{E_0}{K} \left( \frac{d}{5} \right) = E_0 d \left( \frac{4}{5} + \frac{1}{5 \times 6} \right) \).

\( V = E_0 d \left( \frac{4}{5} + \frac{1}{30} \right) = E_0 d \times \frac{25}{30} = E_0 d \times \frac{5}{6} \).

\( C = \frac{Q}{V} = \frac{Q}{\frac{5}{6} V_0} = \frac{6}{5} \times 150 = 180 \, \text{pF} \).

170 pF
175 pF
185 pF
180 pF
4

In an electrostatic shielding scenario, a cavity inside a conductor contains a charge. What can be said about the electric field outside the conductor?

In electrostatic equilibrium, a conductor with a cavity shields the external region from charges inside the cavity. The charge inside induces an equal and opposite charge on the inner surface of the cavity, and the conductor adjusts its outer surface charge distribution to ensure the field inside the conductor (outside the cavity) is zero. The field outside depends only on the net charge on the conductor's outer surface, which includes the induced charges due to the internal charge but is not shielded, so the field outside is non-zero and unaffected by the cavity's internal configuration.

It is zero due to shielding
It depends only on the conductor's shape
It depends on the net charge on the conductor
It is zero regardless of the charge on the conductor
2

Two charges \( 18 \, \mu\text{C} \) and \( -6 \, \mu\text{C} \) are placed 18 cm apart. What is the potential energy of the system? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

\( U = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r} = 9 \times 10^9 \times \frac{18 \times 10^{-6} \times (-6 \times 10^{-6})}{0.18} \).

\( U = 9 \times 10^9 \times \frac{-108 \times 10^{-12}}{0.18} = -5.4 \, \text{J} \).

-5 J
-5.5 J
-5.4 J
-6 J
3

A parallel plate capacitor has plates of area \( 0.05 \, \text{m}^2 \) separated by 1 mm in air. What is its capacitance? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

\( C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 0.05}{10^{-3}} = 4.425 \times 10^{-10} \, \text{F} = 442.5 \, \text{pF} \).

442.5 pF
400 pF
500 pF
600 pF
1

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