Electrostatic Potential and Capacitance Chapter-Wise Test 16

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A point charge \( Q = 15 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 5 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: \( V = 9 \times 10^9 \times \frac{15 \times 10^{-9}}{5} = 9 \times 10^9 \times 3 \times 10^{-9} = 27 \, \text{V} \).

27 V
30 V
24 V
20 V
1

A charge of \( 3 \, \mu\text{C} \) is brought from infinity to a point where the potential is \( 400 \, \text{V} \). What is the work done?

Work done = Potential energy = \( q V \).

\( W = 3 \times 10^{-6} \times 400 = 1.2 \times 10^{-3} \, \text{J} = 1.2 \, \text{mJ} \).

1.2 mJ
1.5 mJ
1 mJ
2 mJ
1

A parallel plate capacitor with \( C = 40 \, \text{pF} \) in air has a dielectric (\( K = 5 \)) inserted fully between plates. What is the new capacitance?

\( C' = K C = 5 \times 40 = 200 \, \text{pF} \).

160 pF
200 pF
240 pF
280 pF
2

A conductor has a surface charge density of \( 5.5 \times 10^{-6} \, \text{C/m}^2 \). What is the electric field just outside it? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

\( E = \frac{\sigma}{\varepsilon_0} = \frac{5.5 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 6.215 \times 10^5 \, \text{N/C} \).

5.5 × 10⁵ N/C
6.215 × 10⁵ N/C
6.5 × 10⁵ N/C
7 × 10⁵ N/C
2

A charge of \( 6 \, \mu\text{C} \) is moved from infinity to a point where the potential is \( 150 \, \text{V} \). What is the work done?

Work done = Potential energy = \( q V \).

\( W = 6 \times 10^{-6} \times 150 = 9 \times 10^{-4} \, \text{J} = 0.9 \, \text{mJ} \).

0.9 mJ
1 mJ
0.8 mJ
1.2 mJ
1

A \( 3 \, \mu\text{F} \) capacitor is charged to \( 300 \, \text{V} \). What is the energy stored in it?

\( U = \frac{1}{2} C V^2 = \frac{1}{2} \times 3 \times 10^{-6} \times (300)^2 = \frac{1}{2} \times 3 \times 10^{-6} \times 9 \times 10^4 = 0.135 \, \text{J} \).

0.1 J
0.15 J
0.135 J
0.2 J
3

Three charges \( +3 \, \mu\text{C} \), \( -2 \, \mu\text{C} \), and \( +4 \, \mu\text{C} \) are at \( (0, 0, 0) \), \( (3, 0, 0) \), and \( (0, 4, 0) \, \text{m} \). What is the potential at \( (3, 4, 0) \, \text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Distances: \( r_1 = \sqrt{(3-0)^2 + (4-0)^2} = 5 \, \text{m} \), \( r_2 = 4 \, \text{m} \), \( r_3 = 3 \, \text{m} \).

\( V = 9 \times 10^9 \left( \frac{3 \times 10^{-6}}{5} + \frac{-2 \times 10^{-6}}{4} + \frac{4 \times 10^{-6}}{3} \right) \).

\( V = 9 \times 10^9 \left( 0.6 \times 10^{-6} - 0.5 \times 10^{-6} + 1.33 \times 10^{-6} \right) = 9 \times 10^9 \times 1.43 \times 10^{-6} = 1.287 \times 10^4 \, \text{V} \approx 12870 \, \text{V} \).

12000 V
12870 V
13000 V
14000 V
2

A dipole \( p = 7 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 0^\circ \) in a field \( E = 2 \times 10^5 \, \text{N/C} \). What is the work done?

Work done: \( W = p E (\cos \theta_0 - \cos \theta_1) = 7 \times 10^{-9} \times 2 \times 10^5 \times (\cos 90^\circ - \cos 0^\circ) \).

\( W = 7 \times 10^{-9} \times 2 \times 10^5 \times (0 - 1) = -1.4 \times 10^{-3} \, \text{J} \).

-1.2 × 10⁻³ J
-1.5 × 10⁻³ J
-1.6 × 10⁻³ J
-1.4 × 10⁻³ J
4

A charge of \( 5 \, \mu\text{C} \) is moved from infinity to a point where the potential is \( 300 \, \text{V} \). What is the work done?

Work done = Potential energy = \( q V \).

\( W = 5 \times 10^{-6} \times 300 = 1.5 \times 10^{-3} \, \text{J} = 1.5 \, \text{mJ} \).

1.5 mJ
1.2 mJ
1.8 mJ
2 mJ
1

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

Distances: \( r_1 = \sqrt{9^2 + 9^2} = 9\sqrt{2} \, \text{m} \), \( r_2 = 9 \, \text{m} \), \( r_3 = 9 \, \text{m} \).

\( V = 9 \times 10^9 \left( \frac{9 \times 10^{-6}}{9\sqrt{2}} + \frac{-6 \times 10^{-6}}{9} + \frac{4 \times 10^{-6}}{9} \right) \).

\( V = 9 \times 10^9 \left( \frac{9 \times 10^{-6}}{12.728} - \frac{6 \times 10^{-6}}{9} + \frac{4 \times 10^{-6}}{9} \right) \).

\( V = 9 \times 10^9 \left( 0.707 \times 10^{-6} - 0.667 \times 10^{-6} + 0.444 \times 10^{-6} \right) \).

\( V = 9 \times 10^9 \times 0.484 \times 10^{-6} = 4356 \, \text{V} \).

4200 V
4356 V
4400 V
4500 V
2

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