Electrostatic Potential and Capacitance Chapter-Wise Test 11

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Why does the capacitance of a parallel plate capacitor increase when the plates are moved closer while maintaining the same dielectric?

The capacitance of a parallel plate capacitor is \( C = \frac{K \varepsilon_0 A}{d} \), where \( d \) is the separation between plates, \( A \) is the area, and \( K \) is the dielectric constant. When the plates are moved closer, \( d \) decreases, and since \( C \propto \frac{1}{d} \), the capacitance increases. A smaller \( d \) means a stronger field for the same charge (\( E = \frac{Q}{\varepsilon_0 A} \)), allowing more charge storage for the same voltage, thus increasing \( C \).

The electric field decreases
The separation decreases, increasing capacitance
The charge on the plates increases
The dielectric constant changes
2

A dipole \( p = 3 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \( 180^\circ \) in a field \( E = 4 \times 10^5 \, \text{N/C} \). What is the work done?

Work done: \( W = p E (\cos \theta_0 - \cos \theta_1) = 3 \times 10^{-9} \times 4 \times 10^5 \times (\cos 0^\circ - \cos 180^\circ) \).

\( W = 3 \times 10^{-9} \times 4 \times 10^5 \times (1 - (-1)) = 3 \times 10^{-9} \times 4 \times 10^5 \times 2 = 2.4 \times 10^{-3} \, \text{J} \).

2 × 10⁻³ J
2.2 × 10⁻³ J
2.6 × 10⁻³ J
2.4 × 10⁻³ J
4

Why does the electric field inside the dielectric of a parallel plate capacitor decrease when the dielectric is inserted while keeping the plates disconnected?

When a dielectric (\( K > 1 \)) is inserted into a disconnected capacitor, the charge \( Q \) remains constant. The dielectric polarizes, creating an induced field opposing the applied field. The effective field inside the dielectric becomes \( E = \frac{E_0}{K} \), where \( E_0 = \frac{\sigma}{\varepsilon_0} \) is the field without the dielectric (\( \sigma = Q/A \)). Since \( K > 1 \), the field decreases due to the polarization reducing the net field between the plates.

Due to increased capacitance
Due to decreased potential difference
Due to increased charge on the plates
Due to polarization opposing the applied field
4

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

\( \frac{1}{C} = \frac{1}{25} + \frac{1}{25} + \frac{1}{25} + \frac{1}{25} = \frac{4}{25} \).

\( C = \frac{25}{4} = 6.25 \, \mu\text{F} \).

6 µF
6.5 µF
7 µF
6.25 µF
4

A \( 7 \, \mu\text{F} \) capacitor is charged to \( 400 \, \text{V} \). What is the energy stored in it?

\( U = \frac{1}{2} C V^2 = \frac{1}{2} \times 7 \times 10^{-6} \times (400)^2 \).

\( U = \frac{1}{2} \times 7 \times 10^{-6} \times 160000 = 0.56 \, \text{J} \).

0.5 J
0.6 J
0.56 J
0.7 J
3

A parallel plate capacitor with \( C = 50 \, \text{pF} \) in air has a dielectric (\( K = 4 \)) inserted fully between plates. What is the new capacitance?

\( C' = K C = 4 \times 50 = 200 \, \text{pF} \).

180 pF
200 pF
220 pF
240 pF
2

Two charges \( 15 \, \mu\text{C} \) and \( -5 \, \mu\text{C} \) are placed 12 cm apart. What is the potential energy of the system? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

\( U = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r} = 9 \times 10^9 \times \frac{15 \times 10^{-6} \times (-5 \times 10^{-6})}{0.12} \).

\( U = 9 \times 10^9 \times \frac{-75 \times 10^{-12}}{0.12} = -5.625 \, \text{J} \).

-5 J
-6 J
-5.625 J
-5.5 J
3

A spherical conductor of radius 10 cm has a charge of \( 2 \times 10^{-7} \, \text{C} \). What is the electric field at a point 15 cm from its center? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

For \( r = 0.15 \, \text{m} > R = 0.1 \, \text{m} \), \( E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2} \).

\( E = 9 \times 10^9 \times \frac{2 \times 10^{-7}}{(0.15)^2} = 9 \times 10^9 \times \frac{2 \times 10^{-7}}{0.0225} = 8 \times 10^4 \, \text{N/C} \).

6 × 10⁴ N/C
7 × 10⁴ N/C
8 × 10⁴ N/C
9 × 10⁴ N/C
3

Two capacitors of \( 20 \, \text{pF} \) and \( 30 \, \text{pF} \) are connected in series. What is the equivalent capacitance?

\( \frac{1}{C} = \frac{1}{20} + \frac{1}{30} = \frac{3 + 2}{60} = \frac{5}{60} \).

\( C = \frac{60}{5} = 12 \, \text{pF} \).

10 pF
12 pF
15 pF
18 pF
2

In a system where a charged conductor is placed inside a larger uncharged hollow conductor, why does the inner conductor induce charges on both the inner and outer surfaces of the outer conductor?

When a charged conductor (charge \( Q \)) is placed inside a larger uncharged hollow conductor, the electric field from the inner charge induces \( -Q \) on the inner surface of the hollow conductor to ensure the field inside the conductor's material is zero. Since the hollow conductor is initially uncharged, its total charge must remain zero, so \( +Q \) appears on its outer surface to balance the induced \( -Q \), maintaining charge conservation and equilibrium.

To increase the potential difference
Due to the dielectric properties of the conductor
To minimize the electric field outside
To satisfy charge conservation and equilibrium
4

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