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  4. Electrostatic Potential and Capacitance Chapter-Wise Test 3

Electrostatic Potential and Capacitance Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Four capacitors of \( 5 \, \mu\text{F} \) each are in series. What is the equivalent capacitance?

\( \frac{1}{C} = \frac{1}{5} + \frac{1}{5} + \frac{1}{5} + \frac{1}{5} = \frac{4}{5} \).

\( C = \frac{5}{4} = 1.25 \, \mu\text{F} \).

1 µF
1.5 µF
2 µF
1.25 µF
4

When a dielectric slab fills only half the space between the plates of a parallel plate capacitor (connected to a battery), why does the electric field differ in the dielectric and air regions?

With a constant voltage \( V \) across the plates, the electric field \( E \) varies between regions. In the air region, \( E_{\text{air}} = \frac{V}{d} \), where \( d \) is the plate separation. In the dielectric region (\( K > 1 \)), polarization reduces the field: \( E_{\text{dielectric}} = \frac{E_{\text{air}}}{K} = \frac{V}{K d} \). Since \( K > 1 \), \( E_{\text{dielectric}} < E_{\text{air}} \). The field differs because the dielectric polarizes, creating an opposing field that reduces the net field within it, while the air region experiences the full field.

Due to non-uniform charge distribution
Due to the dielectric's conductivity
Due to polarization reducing the field in the dielectric
Due to increased capacitance in the air region
1

A dipole \( p = 8 \times 10^{-10} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \( 90^\circ \) in a field \( E = 10^6 \, \text{N/C} \). What is the work done?

Work done: \( W = p E (\cos \theta_0 - \cos \theta_1) = 8 \times 10^{-10} \times 10^6 \times (\cos 0^\circ - \cos 90^\circ) = 8 \times 10^{-4} \times (1 - 0) = 8 \times 10^{-4} \, \text{J} \).

6 × 10⁻⁴ J
7 × 10⁻⁴ J
9 × 10⁻⁴ J
8 × 10⁻⁴ J
4

An electric dipole with moment \( p = 10^{-9} \, \text{C m} \) lies along the y-axis. What is the potential at \( (0, 3, 0) \, \text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

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

\( V = 9 \times 10^9 \times \frac{10^{-9}}{3^2} = 9 \times 10^9 \times \frac{10^{-9}}{9} = 1 \, \text{V} \).

1 V
2 V
3 V
0.5 V
1

A dipole with \( p = 5 \times 10^{-9} \, \text{C m} \) makes an angle of \( 30^\circ \) with a uniform field \( E = 3 \times 10^4 \, \text{N/C} \). What is its potential energy?

\( U = -p E \cos \theta = -5 \times 10^{-9} \times 3 \times 10^4 \times \cos 30^\circ \).

\( \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \), so \( U = -5 \times 10^{-9} \times 3 \times 10^4 \times 0.866 = -1.3 \times 10^{-4} \, \text{J} \).

-1.5 × 10⁻⁴ J
-1.3 × 10⁻⁴ J
-1 × 10⁻⁴ J
-2 × 10⁻⁴ J
2

A spherical conductor of radius 5 cm has a charge of \( 5 \times 10^{-8} \, \text{C} \). What is the potential at its surface? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Potential at the surface: \( V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R} \).

\( V = 9 \times 10^9 \times \frac{5 \times 10^{-8}}{0.05} = 9 \times 10^9 \times 10^{-6} = 9000 \, \text{V} \).

9000 V
8000 V
10000 V
12000 V
1

A \( 6 \, \mu\text{F} \) capacitor charged to \( 150 \, \text{V} \) is connected to an uncharged \( 9 \, \mu\text{F} \) capacitor. What is the energy lost?

Initial energy: \( U_i = \frac{1}{2} \times 6 \times 10^{-6} \times (150)^2 = 0.0675 \, \text{J} \).

Charge: \( Q = 6 \times 10^{-6} \times 150 = 9 \times 10^{-4} \, \text{C} \).

Total \( C = 6 + 9 = 15 \, \mu\text{F} \), \( V = \frac{9 \times 10^{-4}}{15 \times 10^{-6}} = 60 \, \text{V} \).

Final energy: \( U_f = \frac{1}{2} \times 15 \times 10^{-6} \times (60)^2 = 0.027 \, \text{J} \).

Loss: \( U_i - U_f = 0.0675 - 0.027 = 0.0405 \, \text{J} \).

0.03 J
0.035 J
0.045 J
0.0405 J
4

A parallel plate capacitor with \( A = 0.01 \, \text{m}^2 \), \( d = 1 \, \text{mm} \) has a dielectric (\( K = 5 \)). What is its capacitance? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

\( C = \frac{\varepsilon_0 K A}{d} = \frac{8.85 \times 10^{-12} \times 5 \times 0.01}{10^{-3}} = 4.425 \times 10^{-10} \, \text{F} = 442.5 \, \text{pF} \).

400 pF
442.5 pF
450 pF
500 pF
2

Why does the potential due to an electric dipole fall off as \( 1/r^2 \) at large distances, unlike the potential of a point charge which falls off as \( 1/r \)?

The potential due to a point charge falls as \( 1/r \) because it behaves as a single source of charge. An electric dipole consists of two equal and opposite charges separated by a small distance, so their potentials partially cancel out at large distances. The net potential depends on the dipole moment and the angle between the position vector and dipole axis, leading to a \( 1/r^2 \) dependence, as derived from the superposition of potentials from the two charges, with higher-order terms becoming negligible at large \( r \).

The dipole has a larger charge magnitude
The dipole's charges cancel out more rapidly with distance
The dipole's field is always zero at large distances
The dipole has a larger separation between charges
2

A \( 9 \, \mu\text{F} \) capacitor is charged to \( 600 \, \text{V} \). What is the energy stored in it?

\( U = \frac{1}{2} C V^2 = \frac{1}{2} \times 9 \times 10^{-6} \times (600)^2 \).

\( U = \frac{1}{2} \times 9 \times 10^{-6} \times 360000 = 1.62 \, \text{J} \).

1.5 J
1.7 J
1.62 J
1.8 J
3

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