Electrostatic Potential and Capacitance Chapter-Wise Test 17

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A parallel plate capacitor has plates of area \( 0.09 \, \text{m}^2 \) and separation 0.5 mm in air. What is its capacitance? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

\( C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 0.09}{0.5 \times 10^{-3}} = 1.593 \times 10^{-9} \, \text{F} = 1593 \, \text{pF} \).

1593 pF
1500 pF
1600 pF
1700 pF
1

Two charges \( 6 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) are 12 cm apart. What is the potential energy of the system? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

\( U = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r} = 9 \times 10^9 \times \frac{6 \times 10^{-6} \times (-3 \times 10^{-6})}{0.12} = 9 \times 10^9 \times \frac{-18 \times 10^{-12}}{0.12} = -1.35 \, \text{J} \).

-1.5 J
-1.35 J
-1.2 J
-1 J
2

A parallel plate capacitor with \( C = 30 \, \text{pF} \) in air has a dielectric (\( K = 6 \)) inserted fully between plates. What is the new capacitance?

\( C' = K C = 6 \times 30 = 180 \, \text{pF} \).

150 pF
180 pF
200 pF
220 pF
2

Two capacitors of \( 40 \, \text{pF} \) and \( 80 \, \text{pF} \) are connected in series. What is the equivalent capacitance?

\( \frac{1}{C} = \frac{1}{40} + \frac{1}{80} = \frac{2 + 1}{80} = \frac{3}{80} \).

\( C = \frac{80}{3} \approx 26.67 \, \text{pF} \).

25 pF
26.67 pF
30 pF
35 pF
2

A parallel plate capacitor has plates of area \( 0.06 \, \text{m}^2 \) and separation 0.3 mm in air. What is its capacitance? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

\( C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 0.06}{0.3 \times 10^{-3}} = 1.77 \times 10^{-9} \, \text{F} = 1770 \, \text{pF} \).

1770 pF
1500 pF
2000 pF
1000 pF
1

Why does the electric field inside a conductor remain zero when a non-symmetric external field is applied?

In electrostatic equilibrium, the electric field inside a conductor must be zero, regardless of the external field’s symmetry. Free charges redistribute on the conductor’s surface to cancel the external field inside. For a non-symmetric field, the surface charge distribution adjusts accordingly—more charges where the field is stronger and fewer where weaker—ensuring the net field inside is zero, as any internal field would cause charge motion, contradicting equilibrium.

The field cancels due to symmetry
The conductor absorbs the field
The field induces opposite charges
Charge redistribution cancels the field inside
4

A \( 8 \, \mu\text{F} \) capacitor charged to \( 50 \, \text{V} \) is connected to an uncharged \( 8 \, \mu\text{F} \) capacitor. What is the final potential difference?

Initial charge: \( Q = 8 \times 10^{-6} \times 50 = 4 \times 10^{-4} \, \text{C} \).

Total capacitance: \( 8 + 8 = 16 \, \mu\text{F} \).

Final voltage: \( V = \frac{Q}{C} = \frac{4 \times 10^{-4}}{16 \times 10^{-6}} = 25 \, \text{V} \).

20 V
30 V
25 V
40 V
3

A spherical conductor of radius 8 cm has a charge of \( 4 \times 10^{-8} \, \text{C} \). What is the electric field at 12 cm from the center? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

For \( r = 0.12 \, \text{m} > R = 0.08 \, \text{m} \), \( E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2} \).

\( E = 9 \times 10^9 \times \frac{4 \times 10^{-8}}{(0.12)^2} = 9 \times 10^9 \times \frac{4 \times 10^{-8}}{0.0144} = 2.5 \times 10^4 \, \text{N/C} \).

2 × 10⁴ N/C
3 × 10⁴ N/C
2.5 × 10⁴ N/C
4 × 10⁴ N/C
3

Why does the potential at the surface of a charged conductor vary if the conductor has an irregular shape?

For a conductor in electrostatic equilibrium, the potential is constant throughout its volume and surface because \( E = 0 \) inside, so no work is done moving a charge within. However, for an irregular shape, the surface charge density \( \sigma \) varies (higher at sharper regions due to higher curvature), affecting the field just outside (\( E = \frac{\sigma}{\varepsilon_0} \)). Despite this, the potential \( V \) on the surface remains constant across the entire surface, as any variation would cause charge flow, contradicting equilibrium. The question may imply field variation, but potential is constant.

Due to varying electric field strength
It stays constant to maintain equilibrium
Due to non-uniform charge density
Due to irregular field lines
2

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

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

\( V = 9 \times 10^9 \times \frac{2 \times 10^{-9}}{4^2} = 9 \times 10^9 \times \frac{2 \times 10^{-9}}{16} = 1.125 \, \text{V} \).

1.125 V
1 V
1.5 V
2 V
1

Performance Summary

Score:

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