Electrostatic Potential and Capacitance Chapter-Wise Test 20

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A dipole \( p = 9 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 180^\circ \) in a field \( E = 5 \times 10^4 \, \text{N/C} \). What is the work done?

Work done: \( W = p E (\cos \theta_0 - \cos \theta_1) = 9 \times 10^{-9} \times 5 \times 10^4 \times (\cos 90^\circ - \cos 180^\circ) \).

\( W = 9 \times 10^{-9} \times 5 \times 10^4 \times (0 - (-1)) = 9 \times 10^{-9} \times 5 \times 10^4 \times 1 = 4.5 \times 10^{-4} \, \text{J} \).

4 × 10⁻⁴ J
5 × 10⁻⁴ J
4.2 × 10⁻⁴ J
4.5 × 10⁻⁴ J
4

Two charges \( 5 \, \mu\text{C} \) and \( -5 \, \mu\text{C} \) are placed 10 cm apart. What is the potential energy of the system? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

\( U = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r} = 9 \times 10^9 \times \frac{5 \times 10^{-6} \times (-5 \times 10^{-6})}{0.1} = 9 \times 10^9 \times \frac{-25 \times 10^{-12}}{0.1} = -2.25 \, \text{J} \).

-2 J
-2.5 J
-2.25 J
-3 J
3

An electric dipole with moment \( p = 3 \times 10^{-9} \, \text{C m} \) lies along the x-axis. What is the potential at \( (0, 5, 0) \, \text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

In the equatorial plane (\( \theta = 90^\circ \)): \( V = \frac{1}{4 \pi \varepsilon_0} \frac{p \cos \theta}{r^2} \).

Since \( \cos 90^\circ = 0 \), \( V = 0 \, \text{V} \).

0 V
1 V
2 V
0.5 V
1

A charged particle is placed in a region where the electric potential varies linearly with distance. What can be said about the electric field in that region?

The electric field \( E \) is related to potential \( V \) by \( E = -\frac{dV}{dx} \). If \( V \) varies linearly with distance (\( V = kx + c \)), then \( \frac{dV}{dx} = k \), a constant. Thus, \( E = -k \), meaning the electric field is uniform (constant magnitude and direction) in the direction opposite to the gradient of \( V \), consistent with a linear potential gradient creating a constant field.

It varies inversely with distance
It varies linearly with distance
It varies quadratically with distance
It is uniform and constant
4

A spherical conductor of radius 3 cm has a charge of \( 3 \times 10^{-8} \, \text{C} \). 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: \( V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R} \).

\( V = 9 \times 10^9 \times \frac{3 \times 10^{-8}}{0.03} = 9 \times 10^9 \times 10^{-6} = 9000 \, \text{V} \).

9000 V
8000 V
10000 V
12000 V
1

An electric dipole with moment \( p = 7 \times 10^{-9} \, \text{C m} \) lies along the x-axis. What is the potential at \( (4, 0, 0) \, \text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

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

\( V = 9 \times 10^9 \times \frac{7 \times 10^{-9}}{4^2} = 9 \times 10^9 \times \frac{7 \times 10^{-9}}{16} = 3.9375 \, \text{V} \).

3.9375 V
4 V
3.5 V
5 V
1

A parallel plate capacitor with \( C = 80 \, \text{pF} \) in air has a dielectric (\( K = 8 \)) inserted fully between plates. What is the new capacitance?

\( C' = K C = 8 \times 80 = 640 \, \text{pF} \).

600 pF
640 pF
700 pF
800 pF
2

A \( 5 \, \mu\text{F} \) capacitor charged to \( 200 \, \text{V} \) is connected to an uncharged \( 5 \, \mu\text{F} \) capacitor. What is the final potential difference?

Initial charge: \( Q = 5 \times 10^{-6} \times 200 = 10^{-3} \, \text{C} \).

Total capacitance: \( 5 + 5 = 10 \, \mu\text{F} \).

Final voltage: \( V = \frac{Q}{C} = \frac{10^{-3}}{10 \times 10^{-6}} = 100 \, \text{V} \).

80 V
120 V
100 V
150 V
3

A spherical conductor of radius 4 cm has a charge of \( 8 \times 10^{-8} \, \text{C} \). 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: \( V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R} \).

\( V = 9 \times 10^9 \times \frac{8 \times 10^{-8}}{0.04} = 9 \times 10^9 \times 2 \times 10^{-6} = 18000 \, \text{V} \).

18000 V
15000 V
20000 V
12000 V
1

A parallel plate capacitor with capacitance \( 300 \, \text{pF} \) has a dielectric (\( K = 3 \), thickness \( d/8 \)) inserted. What is the new capacitance? (Original separation \( d \)).

Potential difference: \( V = E_0 \left( \frac{7d}{8} \right) + \frac{E_0}{K} \left( \frac{d}{8} \right) = E_0 d \left( \frac{7}{8} + \frac{1}{8 \times 3} \right) \).

\( V = E_0 d \left( \frac{7}{8} + \frac{1}{24} \right) = E_0 d \times \frac{22}{24} = E_0 d \times \frac{11}{12} \).

\( C = \frac{Q}{V} = \frac{Q}{\frac{11}{12} V_0} = \frac{12}{11} \times 300 \approx 327.27 \, \text{pF} \).

320 pF
325 pF
330 pF
327.27 pF
4

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