Electrostatic Potential and Capacitance Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 10 \, \mu\text{F} \) capacitor charged to \( 60 \, \text{V} \) is connected to an uncharged \( 30 \, \mu\text{F} \) capacitor. What is the energy lost?

Initial energy: \( U_i = \frac{1}{2} \times 10 \times 10^{-6} \times (60)^2 = 0.018 \, \text{J} \).

Charge: \( Q = 10 \times 10^{-6} \times 60 = 6 \times 10^{-4} \, \text{C} \).

Total \( C = 10 + 30 = 40 \, \mu\text{F} \), \( V = \frac{6 \times 10^{-4}}{40 \times 10^{-6}} = 15 \, \text{V} \).

Final energy: \( U_f = \frac{1}{2} \times 40 \times 10^{-6} \times (15)^2 = 0.0045 \, \text{J} \).

Loss: \( U_i - U_f = 0.018 - 0.0045 = 0.0135 \, \text{J} \).

0.012 J
0.014 J
0.015 J
0.0135 J
4

A point charge \( Q = 2 \times 10^{-6} \, \text{C} \) is placed at the origin. What is the electrostatic potential at a point \( P \) located at \( (3, 4, 0) \, \text{m} \) from the origin? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Distance from origin to \( P \): \( r = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \, \text{m} \).

Potential due to a point charge: \( V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r} \).

Substitute: \( V = 9 \times 10^9 \times \frac{2 \times 10^{-6}}{5} = 9 \times 10^9 \times 4 \times 10^{-7} = 3.6 \times 10^3 \, \text{V} \).

3600 V
3000 V
4000 V
4500 V
1

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

\( V = 9 \times 10^9 \times \frac{4 \times 10^{-8}}{0.08} = 9 \times 10^9 \times 5 \times 10^{-7} = 4500 \, \text{V} \).

4500 V
4000 V
5000 V
6000 V
1

A dipole with \( p = 2 \times 10^{-9} \, \text{C m} \) makes an angle of \( 45^\circ \) with a uniform field \( E = 4 \times 10^5 \, \text{N/C} \). What is its potential energy?

\( U = -p E \cos \theta = -2 \times 10^{-9} \times 4 \times 10^5 \times \cos 45^\circ \).

\( \cos 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707 \), so \( U = -2 \times 10^{-9} \times 4 \times 10^5 \times 0.707 = -5.656 \times 10^{-4} \, \text{J} \).

-6 × 10⁻⁴ J
-5.656 × 10⁻⁴ J
-5 × 10⁻⁴ J
-4 × 10⁻⁴ J
2

In a system of two identical charged spheres brought close together, why does the potential at the midpoint differ from the sum of their individual potentials?

The potential at a point due to multiple charges is the algebraic sum of the potentials due to each charge (principle of superposition). However, when two charged spheres are close, they influence each other’s charge distribution (mutual interaction), polarizing each other. This alters the effective potential at the midpoint compared to the sum of their individual potentials calculated in isolation, as the fields overlap and the charge distribution deviates from that of isolated spheres.

Due to the spheres' different charges
Due to superposition of electric fields
Due to the spheres' large separation
Due to mutual interaction altering charge distribution
3

A spherical conductor of radius 5 cm is charged with \( 3 \times 10^{-8} \, \text{C} \). 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: \( V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R} \).

\( V = 9 \times 10^9 \times \frac{3 \times 10^{-8}}{0.05} = 9 \times 10^9 \times 6 \times 10^{-7} = 5400 \, \text{V} \).

5400 V
4800 V
6000 V
7200 V
1

A \( 6 \, \mu\text{F} \) capacitor charged to \( 100 \, \text{V} \) is connected to an uncharged \( 6 \, \mu\text{F} \) capacitor. What is the final potential difference?

Initial charge: \( Q = 6 \times 10^{-6} \times 100 = 6 \times 10^{-4} \, \text{C} \).

Total capacitance: \( 6 + 6 = 12 \, \mu\text{F} \).

Final voltage: \( V = \frac{Q}{C} = \frac{6 \times 10^{-4}}{12 \times 10^{-6}} = 50 \, \text{V} \).

40 V
60 V
50 V
75 V
3

Four charges \( +q, -q, +q, -q \) are at the corners of a square of side \( 2 \, \text{m} \) in order. What is the potential at the center? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \), \( q = 1 \, \mu\text{C} \)).

Distance from center to each corner = \( \sqrt{2} \, \text{m} \).

\( V = 9 \times 10^9 \left( \frac{1 \times 10^{-6}}{\sqrt{2}} + \frac{-1 \times 10^{-6}}{\sqrt{2}} + \frac{1 \times 10^{-6}}{\sqrt{2}} + \frac{-1 \times 10^{-6}}{\sqrt{2}} \right) = 0 \, \text{V} \).

0 V
4500 V
6370 V
9000 V
1

In a system where a charged conductor is placed near an uncharged conductor, why does the uncharged conductor experience a net attractive force?

The charged conductor induces charges on the uncharged conductor: opposite charges on the near side and like charges on the far side. The distance between the opposite charges is smaller than between like charges, so the attractive force (proportional to \( 1/r^2 \)) between opposite charges dominates over the repulsive force between like charges. This imbalance results in a net attractive force, even though the uncharged conductor has zero net charge initially.

Due to uniform charge distribution
Due to induced charges creating stronger attraction
Due to repulsion between like charges
Due to the charged conductor's larger size
1

Five capacitors of \( 30 \, \mu\text{F} \) each are in series. What is the equivalent capacitance?

\( \frac{1}{C} = \frac{1}{30} + \frac{1}{30} + \frac{1}{30} + \frac{1}{30} + \frac{1}{30} = \frac{5}{30} \).

\( C = \frac{30}{5} = 6 \, \mu\text{F} \).

5 µF
7 µF
5.5 µF
6 µF
4

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