Electrostatic Potential and Capacitance Chapter-Wise Test 21

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 8 \, \mu\text{F} \) capacitor is charged to \( 500 \, \text{V} \). What is the energy stored in it?

\( U = \frac{1}{2} C V^2 = \frac{1}{2} \times 8 \times 10^{-6} \times (500)^2 \).

\( U = \frac{1}{2} \times 8 \times 10^{-6} \times 250000 = 1 \, \text{J} \).

0.9 J
1.1 J
1 J
1.2 J
3

A \( 1 \, \mu\text{F} \) capacitor is charged to \( 400 \, \text{V} \). What is the energy stored in it?

\( U = \frac{1}{2} C V^2 = \frac{1}{2} \times 1 \times 10^{-6} \times (400)^2 \).

\( U = \frac{1}{2} \times 1 \times 10^{-6} \times 16 \times 10^4 = 0.08 \, \text{J} \).

0.06 J
0.1 J
0.08 J
0.12 J
3

Three capacitors \( 6 \, \text{pF} \), \( 12 \, \text{pF} \), and \( 4 \, \text{pF} \) are in series. What is the equivalent capacitance?

\( \frac{1}{C} = \frac{1}{6} + \frac{1}{12} + \frac{1}{4} = \frac{2}{12} + \frac{1}{12} + \frac{3}{12} = \frac{6}{12} = 0.5 \).

\( C = 2 \, \text{pF} \).

1.5 pF
2 pF
2.5 pF
3 pF
2

A \( 2 \, \mu\text{F} \) capacitor charged to \( 200 \, \text{V} \) loses how much energy when connected to an uncharged \( 3 \, \mu\text{F} \) capacitor?

Initial energy: \( U_i = \frac{1}{2} \times 2 \times 10^{-6} \times (200)^2 = 0.04 \, \text{J} \).

Charge: \( Q = 2 \times 10^{-6} \times 200 = 4 \times 10^{-4} \, \text{C} \).

Total \( C = 2 + 3 = 5 \, \mu\text{F} \), \( V = \frac{4 \times 10^{-4}}{5 \times 10^{-6}} = 80 \, \text{V} \).

Final energy: \( U_f = \frac{1}{2} \times 5 \times 10^{-6} \times (80)^2 = 0.016 \, \text{J} \).

Loss: \( 0.04 - 0.016 = 0.024 \, \text{J} \).

0.02 J
0.024 J
0.03 J
0.04 J
2

Three charges \( +10 \, \mu\text{C} \), \( -7 \, \mu\text{C} \), and \( +5 \, \mu\text{C} \) are at \( (0, 0, 0) \), \( (10, 0, 0) \), and \( (0, 10, 0) \, \text{m} \). What is the potential at \( (10, 10, 0) \, \text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Distances: \( r_1 = \sqrt{10^2 + 10^2} = 10\sqrt{2} \, \text{m} \), \( r_2 = 10 \, \text{m} \), \( r_3 = 10 \, \text{m} \).

\( V = 9 \times 10^9 \left( \frac{10 \times 10^{-6}}{10\sqrt{2}} + \frac{-7 \times 10^{-6}}{10} + \frac{5 \times 10^{-6}}{10} \right) \).

\( V = 9 \times 10^9 \left( \frac{10 \times 10^{-6}}{14.142} - \frac{7 \times 10^{-6}}{10} + \frac{5 \times 10^{-6}}{10} \right) \).

\( V = 9 \times 10^9 \left( 0.707 \times 10^{-6} - 0.7 \times 10^{-6} + 0.5 \times 10^{-6} \right) \).

\( V = 9 \times 10^9 \times 0.507 \times 10^{-6} = 4563 \, \text{V} \).

4500 V
4563 V
4600 V
4700 V
2

Two charges \( 6 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) are at \( (10, 0, 0) \) and \( (-10, 0, 0) \, \text{cm} \). 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 = 0.1 m.

\( V = 9 \times 10^9 \left( \frac{6 \times 10^{-6}}{0.1} + \frac{-3 \times 10^{-6}}{0.1} \right) = 9 \times 10^9 \times \frac{3 \times 10^{-6}}{0.1} = 2.7 \times 10^5 \, \text{V} \).

2 × 10⁵ V
2.5 × 10⁵ V
3 × 10⁵ V
2.7 × 10⁵ V
4

When a dielectric slab is inserted between the plates of a charged parallel plate capacitor (disconnected from the battery), what happens to the energy stored in the capacitor?

When a dielectric slab (with \( K > 1 \)) is inserted into a disconnected capacitor, the charge \( Q \) remains constant. The capacitance increases (\( C' = K C \)), and the potential difference decreases (\( V' = V/K \)). The energy stored is given by \( U = \frac{Q^2}{2C} \), so with increased \( C \), the energy decreases (\( U' = \frac{Q^2}{2 K C} \)), as \( K > 1 \). The energy decrease occurs because work is done by the electric field to polarize the dielectric.

It increases due to higher capacitance
It decreases due to reduced potential difference
It remains the same due to constant charge
It becomes zero as the field vanishes
2

Why does the potential energy of a system of two charges depend only on their separation and not on their orientation in space?

The potential energy between two point charges is \( U = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r} \), where \( r \) is the distance between them. This expression depends only on the magnitude of the separation \( r \), not on the direction or orientation of the line joining the charges in space, because the Coulomb force is isotropic (depends only on distance) and the potential is a scalar quantity derived from integrating the force over distance, independent of spatial orientation.

Due to the charges' alignment with the field
Due to the field's directional dependence
Due to the charges' orbital motion
Due to the isotropic nature of Coulomb's law
4

Three capacitors \( 8 \, \text{pF} \), \( 16 \, \text{pF} \), and \( 32 \, \text{pF} \) are in parallel. What is the total capacitance?

\( C = 8 + 16 + 32 = 56 \, \text{pF} \).

50 pF
60 pF
56 pF
64 pF
3

Why does the electric field inside a charged spherical shell vary linearly with distance from the center when a uniform dielectric fills the interior?

Electrostatic Potential and Capacitance typically assumes no charge inside a spherical shell, leading to \( E = 0 \). However, if misinterpreted as a charged dielectric sphere (common in advanced contexts but not in the PDF), the field varies as \( E \propto r \). Since the PDF context implies an empty shell or uniform shell charge, \( E = 0 \). Assuming a misinterpretation, the correct context is \( E = 0 \), but for a filled dielectric with uniform charge (not in PDF), \( E \) varies linearly. Here, we clarify: \( E = 0 \) inside a shell, so no variation.

Due to the dielectric's polarization
Due to the field's inverse square law
Due to the shell's non-uniform charge
It does not vary, field is zero inside
4

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0