Electrostatic Potential and Capacitance Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Why does the electric field between the plates of a parallel plate capacitor remain unchanged when the area of overlap between the plates is reduced while keeping the charge constant?

The electric field between the plates of a parallel plate capacitor is \( E = \frac{\sigma}{\varepsilon_0} \), where \( \sigma = \frac{Q}{A} \) is the surface charge density, \( Q \) is the charge, and \( A \) is the overlapping area. If the area \( A \) is reduced while \( Q \) is constant, \( \sigma \) increases (\( \sigma = \frac{Q}{A} \)), but in the region of overlap, \( E = \frac{\sigma}{\varepsilon_0} \) depends on \( \sigma \), and adjustments occur such that \( E \) remains uniform between the overlapping plates. However, the field remains \( E = \frac{V}{d} \) in practice if plates are adjusted, but assuming ideal infinite plates approximation, \( E \) is determined solely by \( \sigma \), and in this context, charge redistribution keeps \( E \) unchanged locally.

The potential difference increases proportionally
The capacitance remains constant
The electric field depends on surface charge density
The charge distributes uniformly across the plates
3

A dipole \( p = 6 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \( 90^\circ \) in a field \( E = 3 \times 10^5 \, \text{N/C} \). What is the work done?

Work done: \( W = p E (\cos \theta_0 - \cos \theta_1) = 6 \times 10^{-9} \times 3 \times 10^5 \times (\cos 0^\circ - \cos 90^\circ) \).

\( W = 6 \times 10^{-9} \times 3 \times 10^5 \times (1 - 0) = 1.8 \times 10^{-3} \, \text{J} \).

1.5 × 10⁻³ J
2 × 10⁻³ J
1.6 × 10⁻³ J
1.8 × 10⁻³ J
4

A \( 3 \, \mu\text{F} \) capacitor charged to \( 150 \, \text{V} \) is connected to an uncharged \( 3 \, \mu\text{F} \) capacitor. What is the final potential difference?

Initial charge: \( Q = 3 \times 10^{-6} \times 150 = 4.5 \times 10^{-4} \, \text{C} \).

Total capacitance: \( 3 + 3 = 6 \, \mu\text{F} \).

Final voltage: \( V = \frac{Q}{C} = \frac{4.5 \times 10^{-4}}{6 \times 10^{-6}} = 75 \, \text{V} \).

50 V
100 V
75 V
90 V
3

Five capacitors of \( 25 \, \mu\text{F} \) each are in series. What is the equivalent capacitance?

\( \frac{1}{C} = \frac{1}{25} + \frac{1}{25} + \frac{1}{25} + \frac{1}{25} + \frac{1}{25} = \frac{5}{25} \).

\( C = \frac{25}{5} = 5 \, \mu\text{F} \).

4 µF
6 µF
4.5 µF
5 µF
4

In a system of two identical capacitors in series, why does removing one capacitor while maintaining the same total voltage across the system increase the energy stored?

Two identical capacitors in series (\( C_{\text{eq}} = \frac{C}{2} \)) with total voltage \( V \) each have voltage \( V/2 \), storing energy \( U_{\text{total}} = \frac{1}{2} \frac{C}{2} V^2 = \frac{1}{4} C V^2 \). Removing one, the remaining capacitor has capacitance \( C \) and voltage \( V \), storing \( U = \frac{1}{2} C V^2 \), which is greater (\( \frac{1}{2} C V^2 > \frac{1}{4} C V^2 \)). The energy increases because the same \( V \) across a larger effective capacitance (single \( C \)) stores more energy.

Charge on the remaining capacitor decreases
The electric field decreases
The single capacitor stores more energy for the same voltage
Capacitance decreases, reducing energy
3

A parallel plate capacitor has plates of area \( 0.1 \, \text{m}^2 \) and separation 0.5 mm in air. What is its capacitance? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

\( C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 0.1}{0.5 \times 10^{-3}} = 1.77 \times 10^{-9} \, \text{F} = 1770 \, \text{pF} \).

1770 pF
1500 pF
2000 pF
1000 pF
1

Two charges \( 12 \, \mu\text{C} \) and \( -6 \, \mu\text{C} \) are placed 15 cm apart. What is the potential energy of the system? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

\( U = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r} = 9 \times 10^9 \times \frac{12 \times 10^{-6} \times (-6 \times 10^{-6})}{0.15} \).

\( U = 9 \times 10^9 \times \frac{-72 \times 10^{-12}}{0.15} = -4.32 \, \text{J} \).

-4 J
-4.5 J
-4.32 J
-5 J
3

Two capacitors of \( 15 \, \text{pF} \) and \( 30 \, \text{pF} \) are connected in series. What is the equivalent capacitance?

\( \frac{1}{C} = \frac{1}{15} + \frac{1}{30} = \frac{2 + 1}{30} = \frac{3}{30} \).

\( C = \frac{30}{3} = 10 \, \text{pF} \).

8 pF
10 pF
12 pF
15 pF
2

A parallel plate capacitor has plates of area \( 0.04 \, \text{m}^2 \) and separation 0.2 mm in air. What is its capacitance? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

\( C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 0.04}{0.2 \times 10^{-3}} = 1.77 \times 10^{-9} \, \text{F} = 1770 \, \text{pF} \).

1770 pF
1500 pF
2000 pF
1200 pF
1

A \( 4 \, \mu\text{F} \) capacitor charged to \( 300 \, \text{V} \) is connected to an uncharged \( 8 \, \mu\text{F} \) capacitor. What is the energy lost?

Initial energy: \( U_i = \frac{1}{2} \times 4 \times 10^{-6} \times (300)^2 = 0.18 \, \text{J} \).

Charge: \( Q = 4 \times 10^{-6} \times 300 = 1.2 \times 10^{-3} \, \text{C} \).

Total \( C = 4 + 8 = 12 \, \mu\text{F} \), \( V = \frac{1.2 \times 10^{-3}}{12 \times 10^{-6}} = 100 \, \text{V} \).

Final energy: \( U_f = \frac{1}{2} \times 12 \times 10^{-6} \times (100)^2 = 0.06 \, \text{J} \).

Loss: \( U_i - U_f = 0.18 - 0.06 = 0.12 \, \text{J} \).

0.1 J
0.15 J
0.08 J
0.12 J
4

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