Electrostatic Potential and Capacitance Chapter-Wise Test 7

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Three charges \( +8 \, \mu\text{C} \), \( -5 \, \mu\text{C} \), and \( +3 \, \mu\text{C} \) are at \( (0, 0, 0) \), \( (8, 0, 0) \), and \( (0, 8, 0) \, \text{m} \). What is the potential at \( (8, 8, 0) \, \text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Distances: \( r_1 = \sqrt{8^2 + 8^2} = 8\sqrt{2} \, \text{m} \), \( r_2 = 8 \, \text{m} \), \( r_3 = 8 \, \text{m} \).

\( V = 9 \times 10^9 \left( \frac{8 \times 10^{-6}}{8\sqrt{2}} + \frac{-5 \times 10^{-6}}{8} + \frac{3 \times 10^{-6}}{8} \right) \).

\( V = 9 \times 10^9 \left( \frac{8 \times 10^{-6}}{11.314} - \frac{5 \times 10^{-6}}{8} + \frac{3 \times 10^{-6}}{8} \right) \).

\( V = 9 \times 10^9 \left( 0.707 \times 10^{-6} - 0.625 \times 10^{-6} + 0.375 \times 10^{-6} \right) \).

\( V = 9 \times 10^9 \times 0.457 \times 10^{-6} = 4113 \, \text{V} \).

4000 V
4113 V
4200 V
4300 V
2

In an isolated system of two capacitors initially charged and then connected in parallel with opposite polarities, what happens to the final energy of the system compared to the initial energy?

When two capacitors are connected in parallel with opposite polarities, charges redistribute such that the net charge on the positive plates (and negative plates) adjusts to a new equilibrium. The initial energy stored in the capacitors (\( U_i = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2 \)) is greater than the final energy after connection due to energy losses during charge redistribution. Some energy is lost as heat or electromagnetic radiation during the transient current flow, so the final energy is less than the initial energy.

Final energy is less than initial energy due to heat loss
Final energy equals initial energy as charge is conserved
Final energy is greater due to increased potential
Final energy becomes zero as charges cancel out
1

Two charges \( 10 \, \mu\text{C} \) 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.

\( V = 9 \times 10^9 \left( \frac{10 \times 10^{-6}}{0.08} + \frac{-2 \times 10^{-6}}{0.08} \right) = 9 \times 10^9 \times \frac{8 \times 10^{-6}}{0.08} \).

\( V = 9 \times 10^9 \times \frac{8 \times 10^{-6}}{0.08} = 9 \times 10^5 \, \text{V} \).

8 × 10⁵ V
8.5 × 10⁵ V
9.5 × 10⁵ V
9 × 10⁵ V
4

Why does the potential energy of a dipole in a uniform field increase when rotated from alignment (\( \theta = 0^\circ \)) to perpendicular (\( \theta = 90^\circ \))?

The potential energy of a dipole is \( U = -p E \cos \theta \). At \( \theta = 0^\circ \), \( \cos 0 = 1 \), \( U = -p E \), the minimum. At \( \theta = 90^\circ \), \( \cos 90 = 0 \), \( U = 0 \). From \( \theta = 0^\circ \) to \( 90^\circ \), \( U \) increases from \( -p E \) to 0 (less negative to zero), as the dipole moves against the torque (\( \tau = p E \sin \theta \)) aligning with the field, requiring external work that increases the system's energy.

The field strength increases
The dipole moment increases
The torque becomes zero
Work is done against the aligning torque
4

A \( 16 \, \mu\text{F} \) capacitor charged to \( 10 \, \text{V} \) is connected to an uncharged \( 16 \, \mu\text{F} \) capacitor. What is the final potential difference?

Initial charge: \( Q = 16 \times 10^{-6} \times 10 = 1.6 \times 10^{-4} \, \text{C} \).

Total capacitance: \( 16 + 16 = 32 \, \mu\text{F} \).

Final voltage: \( V = \frac{Q}{C} = \frac{1.6 \times 10^{-4}}{32 \times 10^{-6}} = 5 \, \text{V} \).

4 V
6 V
5 V
8 V
3

A spherical conductor of radius 6 cm has a charge of \( 6 \times 10^{-8} \, \text{C} \). What is the potential at its surface? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Potential at the surface: \( V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R} \).

\( V = 9 \times 10^9 \times \frac{6 \times 10^{-8}}{0.06} = 9 \times 10^9 \times 10^{-6} = 9000 \, \text{V} \).

9000 V
8000 V
10000 V
12000 V
1

A charged conductor is surrounded by a thin concentric hollow conducting shell. If the shell is grounded, what happens to the charge on the inner conductor?

When the outer shell is grounded (potential \( V = 0 \)), the potential on the inner conductor adjusts due to charge redistribution. If the inner conductor has charge \( Q \), the inner surface of the shell induces \( -Q \), and since the shell's potential is zero, the outer surface of the shell acquires \( +Q \). The charge on the inner conductor remains \( Q \), as grounding the shell does not change the charge on the inner conductor but affects the potential and charge distribution on the shell.

It becomes zero
It induces an equal charge on the shell
It remains unchanged
It transfers entirely to the shell
3

Three capacitors \( 10 \, \text{pF} \), \( 20 \, \text{pF} \), and \( 40 \, \text{pF} \) are in parallel. What is the total capacitance?

\( C = 10 + 20 + 40 = 70 \, \text{pF} \).

60 pF
65 pF
70 pF
75 pF
3

In a parallel plate capacitor with a partially inserted dielectric slab while maintaining constant charge, why does the capacitance decrease as the slab is withdrawn?

Capacitance \( C = \frac{Q}{V} \). With constant charge \( Q \), inserting a dielectric (\( K > 1 \)) increases \( C \) by reducing \( V \) (\( C = \frac{K \varepsilon_0 A}{d} \)). As the slab is withdrawn, the effective dielectric constant decreases (less area has \( K \)), lowering \( C \). Since \( Q \) is constant, \( V \) increases (\( V = \frac{Q}{C} \)), reflecting the reduced capacitance due to less dielectric influence.

The electric field increases
The charge on the plates decreases
The potential difference remains constant
The effective dielectric constant decreases
4

A charge of \( 8 \, \mu\text{C} \) is moved from infinity to a point where the potential is \( 50 \, \text{V} \). What is the work done?

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

\( W = 8 \times 10^{-6} \times 50 = 4 \times 10^{-4} \, \text{J} = 0.4 \, \text{mJ} \).

0.4 mJ
0.5 mJ
0.3 mJ
0.6 mJ
1

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