Electrostatic Potential and Capacitance Chapter-Wise Test 19

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A dipole \( p = 10 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \( 90^\circ \) in a field \( E = 4 \times 10^5 \, \text{N/C} \). What is the work done?

Work done: \( W = p E (\cos \theta_0 - \cos \theta_1) = 10 \times 10^{-9} \times 4 \times 10^5 \times (\cos 0^\circ - \cos 90^\circ) \).

\( W = 10 \times 10^{-9} \times 4 \times 10^5 \times (1 - 0) = 4 \times 10^{-3} \, \text{J} \).

3.5 × 10⁻³ J
4.5 × 10⁻³ J
3.8 × 10⁻³ J
4 × 10⁻³ J
4

A \( 5 \, \mu\text{F} \) capacitor charged to \( 120 \, \text{V} \) is connected to an uncharged \( 15 \, \mu\text{F} \) capacitor. What is the energy lost?

Initial energy: \( U_i = \frac{1}{2} \times 5 \times 10^{-6} \times (120)^2 = 0.036 \, \text{J} \).

Charge: \( Q = 5 \times 10^{-6} \times 120 = 6 \times 10^{-4} \, \text{C} \).

Total \( C = 5 + 15 = 20 \, \mu\text{F} \), \( V = \frac{6 \times 10^{-4}}{20 \times 10^{-6}} = 30 \, \text{V} \).

Final energy: \( U_f = \frac{1}{2} \times 20 \times 10^{-6} \times (30)^2 = 0.009 \, \text{J} \).

Loss: \( U_i - U_f = 0.036 - 0.009 = 0.027 \, \text{J} \).

0.02 J
0.025 J
0.03 J
0.027 J
4

Why does the electric field inside a uniformly charged spherical shell vary linearly with distance from the center if a point charge is placed at the center?

Without the point charge, the field inside a uniformly charged spherical shell is zero (Gauss’s law). With a point charge \( Q \) at the center, the field inside the shell is due to the point charge only (\( E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2} \)), which varies as \( \frac{1}{r^2} \), not linearly. The question may imply a misunderstanding; in standard electrostatics (per the PDF), the shell contributes no field inside, and the point charge's field does not vary linearly. Thus, \( E \propto \frac{1}{r^2} \), correcting the premise to standard interpretation.

Due to the shell's charge distribution
Due to the shell's dielectric properties
It varies as \( \frac{1}{r^2} \), not linearly
Due to the point charge's field variation
4

A parallel plate capacitor with \( C = 20 \, \text{pF} \) in air has a dielectric (\( K = 3 \)) inserted fully between plates. What is the new capacitance?

\( C' = K C = 3 \times 20 = 60 \, \text{pF} \).

40 pF
60 pF
80 pF
100 pF
2

A \( 4 \, \mu\text{F} \) capacitor is charged to \( 50 \, \text{V} \) and then connected to an uncharged \( 2 \, \mu\text{F} \) capacitor. What is the final energy?

Initial charge: \( Q = 4 \times 10^{-6} \times 50 = 2 \times 10^{-4} \, \text{C} \).

Total capacitance: \( 4 + 2 = 6 \, \mu\text{F} \).

Final voltage: \( V = \frac{Q}{C} = \frac{2 \times 10^{-4}}{6 \times 10^{-6}} = 33.33 \, \text{V} \).

Final energy: \( U = \frac{1}{2} C V^2 = \frac{1}{2} \times 6 \times 10^{-6} \times (33.33)^2 = 3.33 \times 10^{-3} \, \text{J} \).

3 × 10⁻³ J
3.33 × 10⁻³ J
3.5 × 10⁻³ J
4 × 10⁻³ J
2

In a system where a charged particle moves along a curved path between two points with different potentials, what can be inferred about the work done by the electric field?

The work done by the electric field on a charge \( q \) moving between two points is \( W = q (V_{\text{final}} - V_{\text{initial}}) \), where \( V \) is the potential. Since the electrostatic field is conservative, this work depends only on the potential difference between the points and not on the path taken (straight or curved). Thus, the work done is path-independent and determined solely by the potential difference, regardless of the curved trajectory.

It depends on the length of the path
It varies with the curvature of the path
It depends only on the potential difference
It is zero due to the curved trajectory
3

A dipole \( p = 5 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 180^\circ \) in a field \( E = 2 \times 10^5 \, \text{N/C} \). What is the work done?

Work done: \( W = p E (\cos \theta_0 - \cos \theta_1) = 5 \times 10^{-9} \times 2 \times 10^5 \times (\cos 90^\circ - \cos 180^\circ) \).

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

0.5 × 10⁻³ J
1.5 × 10⁻³ J
0.8 × 10⁻³ J
1 × 10⁻³ J
4

Four capacitors of \( 20 \, \mu\text{F} \) each are in series. What is the equivalent capacitance?

\( \frac{1}{C} = \frac{1}{20} + \frac{1}{20} + \frac{1}{20} + \frac{1}{20} = \frac{4}{20} \).

\( C = \frac{20}{4} = 5 \, \mu\text{F} \).

4 µF
6 µF
4.5 µF
5 µF
4

A parallel plate capacitor has plates of area \( 0.07 \, \text{m}^2 \) and separation 0.35 mm in air. What is its capacitance? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

\( C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 0.07}{0.35 \times 10^{-3}} = 1.77 \times 10^{-9} \, \text{F} = 1770 \, \text{pF} \).

1770 pF
1500 pF
2000 pF
1000 pF
1

A \( 14 \, \mu\text{F} \) capacitor charged to \( 20 \, \text{V} \) is connected to an uncharged \( 14 \, \mu\text{F} \) capacitor. What is the final potential difference?

Initial charge: \( Q = 14 \times 10^{-6} \times 20 = 2.8 \times 10^{-4} \, \text{C} \).

Total capacitance: \( 14 + 14 = 28 \, \mu\text{F} \).

Final voltage: \( V = \frac{Q}{C} = \frac{2.8 \times 10^{-4}}{28 \times 10^{-6}} = 10 \, \text{V} \).

5 V
15 V
10 V
20 V
3

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