Electrostatic Potential and Capacitance Chapter-Wise Test 22

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A point charge \( Q = 18 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 6 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: \( V = 9 \times 10^9 \times \frac{18 \times 10^{-9}}{6} = 9 \times 10^9 \times 3 \times 10^{-9} = 27 \, \text{V} \).

27 V
30 V
24 V
20 V
1

A parallel plate capacitor with \( C = 70 \, \text{pF} \) in air has a dielectric (\( K = 5 \)) inserted fully between plates. What is the new capacitance?

\( C' = K C = 5 \times 70 = 350 \, \text{pF} \).

300 pF
350 pF
400 pF
450 pF
2

A point charge \( Q = 9 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 3 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: \( V = 9 \times 10^9 \times \frac{9 \times 10^{-9}}{3} = 9 \times 10^9 \times 3 \times 10^{-9} = 27 \, \text{V} \).

27 V
30 V
24 V
21 V
1

Three charges \( +11 \, \mu\text{C} \), \( -8 \, \mu\text{C} \), and \( +6 \, \mu\text{C} \) are at \( (0, 0, 0) \), \( (11, 0, 0) \), and \( (0, 11, 0) \, \text{m} \). What is the potential at \( (11, 11, 0) \, \text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Distances: \( r_1 = \sqrt{11^2 + 11^2} = 11\sqrt{2} \, \text{m} \), \( r_2 = 11 \, \text{m} \), \( r_3 = 11 \, \text{m} \).

\( V = 9 \times 10^9 \left( \frac{11 \times 10^{-6}}{11\sqrt{2}} + \frac{-8 \times 10^{-6}}{11} + \frac{6 \times 10^{-6}}{11} \right) \).

\( V = 9 \times 10^9 \left( \frac{11 \times 10^{-6}}{15.556} - \frac{8 \times 10^{-6}}{11} + \frac{6 \times 10^{-6}}{11} \right) \).

\( V = 9 \times 10^9 \left( 0.707 \times 10^{-6} - 0.727 \times 10^{-6} + 0.545 \times 10^{-6} \right) \).

\( V = 9 \times 10^9 \times 0.525 \times 10^{-6} = 4725 \, \text{V} \).

4600 V
4725 V
4800 V
4900 V
2

Three capacitors \( 5 \, \text{pF} \), \( 10 \, \text{pF} \), and \( 20 \, \text{pF} \) are in parallel. What is the total capacitance?

\( C = 5 + 10 + 20 = 35 \, \text{pF} \).

30 pF
40 pF
35 pF
45 pF
3

A dielectric slab is partially inserted between the plates of a parallel plate capacitor while maintaining a constant voltage across the plates. What happens to the electric field between the plates in the region where the dielectric is present?

When a dielectric slab (with dielectric constant \( K > 1 \)) is inserted between the plates of a capacitor with constant voltage \( V \), the electric field \( E \) in the dielectric region decreases. The electric field in a dielectric is given by \( E = \frac{E_0}{K} \), where \( E_0 = \frac{V}{d} \) is the field without the dielectric (\( d \) is the plate separation). Since \( K > 1 \), \( E \) is reduced in the dielectric region compared to the air region, as the dielectric polarizes and induces charges that oppose the applied field.

It decreases due to the dielectric's polarization
It increases due to increased charge on the plates
It remains unchanged due to constant voltage
It becomes zero inside the dielectric
1

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

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

\( \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \), so \( U = -4 \times 10^{-9} \times 5 \times 10^4 \times 0.866 = -1.732 \times 10^{-4} \, \text{J} \).

-2 × 10⁻⁴ J
-1.732 × 10⁻⁴ J
-1.5 × 10⁻⁴ J
-1 × 10⁻⁴ J
2

Five capacitors of \( 20 \, \mu\text{F} \) each are in series. What is the equivalent capacitance?

\( \frac{1}{C} = \frac{1}{20} + \frac{1}{20} + \frac{1}{20} + \frac{1}{20} + \frac{1}{20} = \frac{5}{20} \).

\( C = \frac{20}{5} = 4 \, \mu\text{F} \).

3 µF
5 µF
2 µF
4 µF
4

A dipole \( p = 2 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 0^\circ \) in a field \( E = 5 \times 10^5 \, \text{N/C} \). What is the work done?

Work done: \( W = p E (\cos \theta_0 - \cos \theta_1) = 2 \times 10^{-9} \times 5 \times 10^5 \times (\cos 90^\circ - \cos 0^\circ) = 2 \times 10^{-9} \times 5 \times 10^5 \times (0 - 1) = -10^{-3} \, \text{J} \).

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

In a system of two identical conductors initially charged differently and then connected by a wire, why does the final energy of the system decrease?

Initially, the conductors have charges \( Q_1 \) and \( Q_2 \), with energy \( U_i = \frac{Q_1^2}{2C} + \frac{Q_2^2}{2C} \). Upon connection, charge redistributes to equal potentials, total charge \( Q_1 + Q_2 \) splits equally (\( \frac{Q_1 + Q_2}{2} \) each), so final energy \( U_f = 2 \times \frac{\left(\frac{Q_1 + Q_2}{2}\right)^2}{2C} = \frac{(Q_1 + Q_2)^2}{4C} \). Typically, \( (Q_1 + Q_2)^2 < Q_1^2 + Q_2^2 \) unless \( Q_1 = Q_2 \), so \( U_f < U_i \), with energy lost as heat or radiation during charge flow.

The conductors lose charge to ground
The potentials become unequal
The capacitance increases
Energy is lost during charge redistribution
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