Electrostatic Potential And Capacitance Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

An electric dipole with moment \( p = 8 \times 10^{-9} \, \text{C m} \) lies along the y-axis. What is the potential at \( (0, -4, 0) \, \text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Along the dipole axis (\( \theta = 180^\circ \)): \( V = -\frac{1}{4 \pi \varepsilon_0} \frac{p}{r^2} \).

\( V = -9 \times 10^9 \times \frac{8 \times 10^{-9}}{4^2} = -9 \times 10^9 \times \frac{8 \times 10^{-9}}{16} = -4.5 \, \text{V} \).

-4.5 V
4.5 V
0 V
-5 V
1

A \( 2 \, \mu\text{F} \) capacitor charged to \( 400 \, \text{V} \) is connected to an uncharged \( 6 \, \mu\text{F} \) capacitor. What is the energy lost?

Initial energy: \( U_i = \frac{1}{2} \times 2 \times 10^{-6} \times (400)^2 = 0.16 \, \text{J} \).

Charge: \( Q = 2 \times 10^{-6} \times 400 = 8 \times 10^{-4} \, \text{C} \).

Total \( C = 2 + 6 = 8 \, \mu\text{F} \), \( V = \frac{8 \times 10^{-4}}{8 \times 10^{-6}} = 100 \, \text{V} \).

Final energy: \( U_f = \frac{1}{2} \times 8 \times 10^{-6} \times (100)^2 = 0.04 \, \text{J} \).

Loss: \( U_i - U_f = 0.16 - 0.04 = 0.12 \, \text{J} \).

0.1 J
0.15 J
0.08 J
0.12 J
4

Why does the energy density in an electric field depend on the square of the field strength?

The energy density in an electric field is given by \( u = \frac{1}{2} \varepsilon_0 E^2 \). This arises from the energy stored in a capacitor (\( U = \frac{1}{2} C V^2 \)), scaled over the volume. For a parallel plate capacitor, \( E = \frac{V}{d} \), \( C = \frac{\varepsilon_0 A}{d} \), so \( U = \frac{1}{2} \frac{\varepsilon_0 A}{d} (E d)^2 = \frac{1}{2} \varepsilon_0 E^2 (A d) \), where \( A d \) is the volume. Thus, energy density \( u = \frac{U}{A d} = \frac{1}{2} \varepsilon_0 E^2 \), showing the \( E^2 \) dependence due to the quadratic relation between energy and field.

Due to the field's linear variation
Due to the permittivity's dependence on \( E \)
Due to the quadratic nature of energy storage
Due to the inverse square law of fields
3

Why does the electric field near the edge of a charged conducting plate differ from the field at the center of the plate?

Near the center of a large charged conducting plate, the field is approximately uniform (\( E = \frac{\sigma}{\varepsilon_0} \)), as the plate behaves like an infinite sheet. At the edges, the field lines fringe outward due to the finite size of the plate, leading to a non-uniform field. The charge density \( \sigma \) may also increase near edges due to higher curvature or boundary effects, causing the field to be stronger and non-perpendicular, differing from the uniform field at the center.

Charge density is uniform across the plate
The field lines are denser at the center
The plate's thickness varies at the edges
Field lines fringe outward at the edges
4

Two charges \( 16 \, \mu\text{C} \) and \( -4 \, \mu\text{C} \) are at \( (4, 0, 0) \) and \( (-4, 0, 0) \, \text{cm} \). 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 = 0.04 m.

\( V = 9 \times 10^9 \left( \frac{16 \times 10^{-6}}{0.04} + \frac{-4 \times 10^{-6}}{0.04} \right) = 9 \times 10^9 \times \frac{12 \times 10^{-6}}{0.04} \).

\( V = 9 \times 10^9 \times \frac{12 \times 10^{-6}}{0.04} = 2.7 \times 10^6 \, \text{V} \).

2.5 × 10⁶ V
2.6 × 10⁶ V
2.8 × 10⁶ V
2.7 × 10⁶ V
4

Two charges \( 32 \, \mu\text{C} \) and \( -16 \, \mu\text{C} \) are placed 32 cm apart. What is the potential energy of the system? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

\( U = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r} = 9 \times 10^9 \times \frac{32 \times 10^{-6} \times (-16 \times 10^{-6})}{0.32} \).

\( U = 9 \times 10^9 \times \frac{-512 \times 10^{-12}}{0.32} = -14.4 \, \text{J} \).

-14 J
-15 J
-14.4 J
-16 J
3

Two charges \( 8 \, \mu\text{C} \) and \( -4 \, \mu\text{C} \) are at \( (-6, 0, 0) \) and \( (6, 0, 0) \, \text{cm} \) in an external field \( E = 10^5/r^2 \, \text{N/C} \). What is the total potential energy? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Mutual energy: \( U_{12} = 9 \times 10^9 \times \frac{8 \times 10^{-6} \times (-4 \times 10^{-6})}{0.12} = -2.4 \, \text{J} \).

External potential: \( V(r) = \frac{10^5}{r} \), at \( r = 0.06 \, \text{m} \), \( V = \frac{10^5}{0.06} = 1.67 \times 10^6 \, \text{V} \).

External energy: \( 8 \times 10^{-6} \times 1.67 \times 10^6 + (-4 \times 10^{-6}) \times 1.67 \times 10^6 = 13.36 - 6.68 = 6.68 \, \text{J} \).

Total: \( -2.4 + 6.68 = 4.28 \, \text{J} \).

4 J
4.28 J
4.5 J
5 J
2

A parallel plate capacitor with \( C = 90 \, \text{pF} \) in air has a dielectric (\( K = 9 \)) inserted fully between plates. What is the new capacitance?

\( C' = K C = 9 \times 90 = 810 \, \text{pF} \).

800 pF
810 pF
820 pF
900 pF
2

A charge of \( 7 \, \mu\text{C} \) is moved from infinity to a point where the potential is \( 100 \, \text{V} \). What is the work done?

Work done = Potential energy = \( q V \).

\( W = 7 \times 10^{-6} \times 100 = 7 \times 10^{-4} \, \text{J} = 0.7 \, \text{mJ} \).

0.7 mJ
0.8 mJ
0.6 mJ
0.9 mJ
1

A point charge \( Q = 5 \times 10^{-9} \, \text{C} \) is placed at the origin. What is the potential at a point 10 m away from the charge? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

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

Substitute: \( V = 9 \times 10^9 \times \frac{5 \times 10^{-9}}{10} = 9 \times 10^9 \times 0.5 \times 10^{-9} = 4.5 \, \text{V} \).

4.5 V
5 V
6 V
7 V
1

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