Electrostatic Potential and Capacitance Chapter-Wise Test 5

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A parallel plate capacitor has plates of area \( 0.03 \, \text{m}^2 \) and separation 0.5 mm in air. What is its capacitance? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

\( C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 0.03}{0.5 \times 10^{-3}} = 5.31 \times 10^{-10} \, \text{F} = 531 \, \text{pF} \).

531 pF
500 pF
600 pF
450 pF
1

A parallel plate capacitor with capacitance \( 200 \, \text{pF} \) has a dielectric (\( K = 5 \), thickness \( d/6 \)) inserted. What is the new capacitance? (Original separation \( d \)).

Potential difference: \( V = E_0 \left( \frac{5d}{6} \right) + \frac{E_0}{K} \left( \frac{d}{6} \right) = E_0 d \left( \frac{5}{6} + \frac{1}{6 \times 5} \right) \).

\( V = E_0 d \left( \frac{5}{6} + \frac{1}{30} \right) = E_0 d \times \frac{26}{30} = E_0 d \times \frac{13}{15} \).

\( C = \frac{Q}{V} = \frac{Q}{\frac{13}{15} V_0} = \frac{15}{13} \times 200 \approx 230.77 \, \text{pF} \).

220 pF
225 pF
235 pF
230.77 pF
4

Two charges \( 7 \, \mu\text{C} \) and \( -2 \, \mu\text{C} \) are at \( (8, 0, 0) \) and \( (-8, 0, 0) \, \text{cm} \). 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 = 0.08 m.

\( V = 9 \times 10^9 \left( \frac{7 \times 10^{-6}}{0.08} + \frac{-2 \times 10^{-6}}{0.08} \right) = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{0.08} = 5.625 \times 10^5 \, \text{V} \).

5 × 10⁵ V
6 × 10⁵ V
5.5 × 10⁵ V
5.625 × 10⁵ V
4

Two capacitors of \( 50 \, \text{pF} \) and \( 100 \, \text{pF} \) are connected in series. What is the equivalent capacitance?

\( \frac{1}{C} = \frac{1}{50} + \frac{1}{100} = \frac{2 + 1}{100} = \frac{3}{100} \).

\( C = \frac{100}{3} \approx 33.33 \, \text{pF} \).

30 pF
33.33 pF
35 pF
40 pF
2

A parallel plate capacitor with \( C = 60 \, \text{pF} \) in air has a dielectric (\( K = 7 \)) inserted fully between plates. What is the new capacitance?

\( C' = K C = 7 \times 60 = 420 \, \text{pF} \).

400 pF
420 pF
450 pF
480 pF
2

In a system where a charged conductor is grounded while placed near a point charge, why does the potential of the conductor become zero?

Grounding a conductor sets its potential to zero by connecting it to Earth, which acts as a reservoir of infinite charge at \( V = 0 \). Charges flow to or from Earth until the conductor's potential is zero. The nearby point charge induces charges on the conductor, but grounding ensures \( V = 0 \), adjusting the conductor's charge to counteract the point charge's potential influence, maintaining \( V = 0 \) at the conductor.

The point charge neutralizes the conductor
The conductor absorbs the point charge's field
Grounding sets the potential to zero
The conductor's charge becomes infinite
3

A \( 4 \, \mu\text{F} \) capacitor charged to \( 150 \, \text{V} \) is connected to an uncharged \( 12 \, \mu\text{F} \) capacitor. What is the energy lost?

Initial energy: \( U_i = \frac{1}{2} \times 4 \times 10^{-6} \times (150)^2 = 0.045 \, \text{J} \).

Charge: \( Q = 4 \times 10^{-6} \times 150 = 6 \times 10^{-4} \, \text{C} \).

Total \( C = 4 + 12 = 16 \, \mu\text{F} \), \( V = \frac{6 \times 10^{-4}}{16 \times 10^{-6}} = 37.5 \, \text{V} \).

Final energy: \( U_f = \frac{1}{2} \times 16 \times 10^{-6} \times (37.5)^2 = 0.01125 \, \text{J} \).

Loss: \( U_i - U_f = 0.045 - 0.01125 = 0.03375 \, \text{J} \).

0.03 J
0.035 J
0.04 J
0.03375 J
4

Three capacitors \( 6 \, \text{pF} \), \( 12 \, \text{pF} \), and \( 24 \, \text{pF} \) are in parallel. What is the total capacitance?

\( C = 6 + 12 + 24 = 42 \, \text{pF} \).

40 pF
45 pF
42 pF
50 pF
3

A \( 8 \, \mu\text{F} \) capacitor charged to \( 80 \, \text{V} \) is connected to an uncharged \( 24 \, \mu\text{F} \) capacitor. What is the energy lost?

Initial energy: \( U_i = \frac{1}{2} \times 8 \times 10^{-6} \times (80)^2 = 0.0256 \, \text{J} \).

Charge: \( Q = 8 \times 10^{-6} \times 80 = 6.4 \times 10^{-4} \, \text{C} \).

Total \( C = 8 + 24 = 32 \, \mu\text{F} \), \( V = \frac{6.4 \times 10^{-4}}{32 \times 10^{-6}} = 20 \, \text{V} \).

Final energy: \( U_f = \frac{1}{2} \times 32 \times 10^{-6} \times (20)^2 = 0.0064 \, \text{J} \).

Loss: \( U_i - U_f = 0.0256 - 0.0064 = 0.0192 \, \text{J} \).

0.015 J
0.02 J
0.018 J
0.0192 J
4

A \( 6 \, \mu\text{F} \) capacitor is charged to \( 300 \, \text{V} \). What is the energy stored in it?

\( U = \frac{1}{2} C V^2 = \frac{1}{2} \times 6 \times 10^{-6} \times (300)^2 \).

\( U = \frac{1}{2} \times 6 \times 10^{-6} \times 90000 = 0.27 \, \text{J} \).

0.2 J
0.3 J
0.27 J
0.35 J
3

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