Electrostatic Potential and Capacitance Chapter-Wise Test 13

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Two capacitors of \( 70 \, \text{pF} \) and \( 140 \, \text{pF} \) are connected in series. What is the equivalent capacitance?

\( \frac{1}{C} = \frac{1}{70} + \frac{1}{140} = \frac{2 + 1}{140} = \frac{3}{140} \).

\( C = \frac{140}{3} \approx 46.67 \, \text{pF} \).

45 pF
46.67 pF
50 pF
55 pF
2

Two charges \( 12 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) are at \( (5, 0, 0) \) and \( (-5, 0, 0) \, \text{cm} \). What is the potential at midpoint? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Distance to midpoint = 0.05 m.

\( V = 9 \times 10^9 \left( \frac{12 \times 10^{-6}}{0.05} + \frac{-3 \times 10^{-6}}{0.05} \right) = 9 \times 10^9 \times \frac{9 \times 10^{-6}}{0.05} \).

\( V = 9 \times 10^9 \times \frac{9 \times 10^{-6}}{0.05} = 1.62 \times 10^6 \, \text{V} \).

1.5 × 10⁶ V
1.6 × 10⁶ V
1.7 × 10⁶ V
1.62 × 10⁶ V
4

A dipole with \( p = 5 \times 10^{-9} \, \text{C m} \) makes an angle of \( 30^\circ \) with a uniform field \( E = 3 \times 10^5 \, \text{N/C} \). What is its potential energy?

\( U = -p E \cos \theta = -5 \times 10^{-9} \times 3 \times 10^5 \times \cos 30^\circ \).

\( \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \), so \( U = -5 \times 10^{-9} \times 3 \times 10^5 \times 0.866 = -1.299 \times 10^{-3} \, \text{J} \).

-1.5 × 10⁻³ J
-1.299 × 10⁻³ J
-1.2 × 10⁻³ J
-1 × 10⁻³ J
2

A dipole \( p = 8 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \( 180^\circ \) in a field \( E = 1 \times 10^5 \, \text{N/C} \). What is the work done?

Work done: \( W = p E (\cos \theta_0 - \cos \theta_1) = 8 \times 10^{-9} \times 1 \times 10^5 \times (\cos 0^\circ - \cos 180^\circ) \).

\( W = 8 \times 10^{-9} \times 1 \times 10^5 \times (1 - (-1)) = 8 \times 10^{-9} \times 1 \times 10^5 \times 2 = 1.6 \times 10^{-3} \, \text{J} \).

1.5 × 10⁻³ J
1.8 × 10⁻³ J
1.4 × 10⁻³ J
1.6 × 10⁻³ J
4

Why does the potential difference between the plates of a parallel plate capacitor remain constant when a dielectric slab is inserted while the capacitor is connected to a battery?

When a capacitor is connected to a battery, the potential difference \( V \) across its plates is fixed by the battery. Inserting a dielectric slab (with \( K > 1 \)) increases the capacitance (\( C' = K C \)), but the battery maintains \( V \). To keep \( V \) constant (\( Q = C V \)), the charge \( Q \) on the plates increases (\( Q' = C' V = K C V \)), as the battery supplies additional charge. Thus, the potential difference remains constant because the battery ensures it.

The dielectric reduces the electric field
The battery maintains a fixed voltage
The charge on the plates remains constant
The capacitance decreases proportionally
2

An isolated charged capacitor is connected across a resistor momentarily. Why does the energy stored in the capacitor decrease to zero after a long time?

When a charged capacitor is connected across a resistor, it discharges through the resistor, forming an RC circuit. The charge \( Q \) on the capacitor decreases exponentially (\( Q(t) = Q_0 e^{-t/RC} \)). After a long time (\( t \to \infty \)), \( Q \to 0 \), so the energy \( U = \frac{Q^2}{2C} \to 0 \). The energy is dissipated as heat in the resistor during discharge, leaving no energy stored in the capacitor.

The capacitor loses its capacitance
The electric field becomes non-uniform
The charge dissipates as heat in the resistor
The resistor absorbs the electric field
2

Three capacitors \( 4 \, \text{pF} \), \( 8 \, \text{pF} \), and \( 16 \, \text{pF} \) are in parallel. What is the total capacitance?

\( C = 4 + 8 + 16 = 28 \, \text{pF} \).

24 pF
26 pF
28 pF
30 pF
3

A \( 6 \, \mu\text{F} \) capacitor charged to \( 100 \, \text{V} \) is connected to an uncharged \( 18 \, \mu\text{F} \) capacitor. What is the energy lost?

Initial energy: \( U_i = \frac{1}{2} \times 6 \times 10^{-6} \times (100)^2 = 0.03 \, \text{J} \).

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

Total \( C = 6 + 18 = 24 \, \mu\text{F} \), \( V = \frac{6 \times 10^{-4}}{24 \times 10^{-6}} = 25 \, \text{V} \).

Final energy: \( U_f = \frac{1}{2} \times 24 \times 10^{-6} \times (25)^2 = 0.0075 \, \text{J} \).

Loss: \( U_i - U_f = 0.03 - 0.0075 = 0.0225 \, \text{J} \).

0.02 J
0.025 J
0.03 J
0.0225 J
4

Three capacitors \( 10 \, \text{pF} \), \( 20 \, \text{pF} \), and \( 40 \, \text{pF} \) are in parallel. What is the total capacitance?

\( C = 10 + 20 + 40 = 70 \, \text{pF} \).

60 pF
65 pF
70 pF
75 pF
3

Two charges \( 18 \, \mu\text{C} \) and \( -9 \, \mu\text{C} \) are at \( (2, 0, 0) \) and \( (-2, 0, 0) \, \text{cm} \). What is the potential at midpoint? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Distance to midpoint = 0.02 m.

\( V = 9 \times 10^9 \left( \frac{18 \times 10^{-6}}{0.02} + \frac{-9 \times 10^{-6}}{0.02} \right) = 9 \times 10^9 \times \frac{9 \times 10^{-6}}{0.02} \).

\( V = 9 \times 10^9 \times \frac{9 \times 10^{-6}}{0.02} = 4.05 \times 10^6 \, \text{V} \).

4 × 10⁶ V
4.1 × 10⁶ V
4.2 × 10⁶ V
4.05 × 10⁶ V
4

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