Electrostatic Potential and Capacitance Chapter-Wise Test 18

Correct answer Carries: 4.

Wrong Answer Carries: -1.

In a system where a positively charged sphere is enclosed by an uncharged conducting shell, why does the outer surface of the shell acquire a negative charge if a negative charge is brought near it?

The positively charged sphere induces a negative charge on the inner surface of the shell and a positive charge on its outer surface to maintain zero field inside the conductor. When a negative charge is brought near the shell, it repels negative charges to the far side of the shell and attracts positive charges closer, altering the charge distribution. The shell, being uncharged initially, redistributes charges so the outer surface gains a net negative charge to balance the influence of the external negative charge while maintaining equilibrium.

Due to the shell's dielectric properties
Due to repulsion by the external negative charge
Due to the inner sphere's positive charge
Due to grounding of the shell
2

A conductor has a surface charge density of \( 1.5 \times 10^{-6} \, \text{C/m}^2 \). What is the electric field just outside it? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

\( E = \frac{\sigma}{\varepsilon_0} = \frac{1.5 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 1.695 \times 10^5 \, \text{N/C} \).

1.5 × 10⁵ N/C
1.695 × 10⁵ N/C
1.8 × 10⁵ N/C
2 × 10⁵ N/C
2

A charge of \( 2 \, \mu\text{C} \) is moved from infinity to a point with potential \( 500 \, \text{V} \). What is the work done?

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

\( W = 2 \times 10^{-6} \times 500 = 10^{-3} \, \text{J} = 1 \, \text{mJ} \).

1 mJ
2 mJ
0.5 mJ
1.5 mJ
1

Two capacitors of \( 30 \, \text{pF} \) and \( 60 \, \text{pF} \) are connected in series. What is the equivalent capacitance?

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

\( C = \frac{60}{3} = 20 \, \text{pF} \).

15 pF
20 pF
25 pF
30 pF
2

Two charges \( 14 \, \mu\text{C} \) and \( -6 \, \mu\text{C} \) are at \( (3, 0, 0) \) and \( (-3, 0, 0) \, \text{cm} \). 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 = 0.03 m.

\( V = 9 \times 10^9 \left( \frac{14 \times 10^{-6}}{0.03} + \frac{-6 \times 10^{-6}}{0.03} \right) = 9 \times 10^9 \times \frac{8 \times 10^{-6}}{0.03} \).

\( V = 9 \times 10^9 \times \frac{8 \times 10^{-6}}{0.03} = 2.4 \times 10^6 \, \text{V} \).

2.2 × 10⁶ V
2.3 × 10⁶ V
2.5 × 10⁶ V
2.4 × 10⁶ V
4

A conductor has a surface charge density of \( 4.5 \times 10^{-6} \, \text{C/m}^2 \). What is the electric field just outside it? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

\( E = \frac{\sigma}{\varepsilon_0} = \frac{4.5 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 5.085 \times 10^5 \, \text{N/C} \).

4.5 × 10⁵ N/C
5.085 × 10⁵ N/C
5.5 × 10⁵ N/C
6 × 10⁵ N/C
2

A \( 3 \, \mu\text{F} \) capacitor charged to \( 200 \, \text{V} \) is connected to an uncharged \( 9 \, \mu\text{F} \) capacitor. What is the energy lost?

Initial energy: \( U_i = \frac{1}{2} \times 3 \times 10^{-6} \times (200)^2 = 0.06 \, \text{J} \).

Charge: \( Q = 3 \times 10^{-6} \times 200 = 6 \times 10^{-4} \, \text{C} \).

Total \( C = 3 + 9 = 12 \, \mu\text{F} \), \( V = \frac{6 \times 10^{-4}}{12 \times 10^{-6}} = 50 \, \text{V} \).

Final energy: \( U_f = \frac{1}{2} \times 12 \times 10^{-6} \times (50)^2 = 0.015 \, \text{J} \).

Loss: \( U_i - U_f = 0.06 - 0.015 = 0.045 \, \text{J} \).

0.03 J
0.04 J
0.05 J
0.045 J
4

When a charged conductor is placed in contact with an uncharged conductor of smaller size, why does the smaller conductor acquire a higher surface charge density upon reaching equilibrium?

When two conductors reach equilibrium, they share charge and attain the same potential. Potential on a conductor's surface is \( V = \frac{Q}{4 \pi \varepsilon_0 R} \) for a sphere (or similar for other shapes). For equal \( V \), \( \frac{Q_1}{R_1} = \frac{Q_2}{R_2} \), so \( Q \propto R \). Surface charge density \( \sigma = \frac{Q}{4 \pi R^2} \), so \( \sigma \propto \frac{Q}{R^2} \propto \frac{R}{R^2} = \frac{1}{R} \). The smaller conductor (smaller \( R \)) has higher \( \sigma \) because \( \sigma \) is inversely proportional to radius.

It has a larger surface area
It stores more charge
It has a smaller radius, increasing density
It has a higher potential
2

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

Position vector \( \mathbf{r} = (0, 2, 0) \), \( r = 2 \, \text{m} \), \( \hat{\mathbf{r}} = (0, 1, 0) \).

Dipole moment \( \mathbf{p} = (2 \times 10^{-10}, 0, 0) \).

\( V = \frac{1}{4 \pi \varepsilon_0} \frac{\mathbf{p} \cdot \hat{\mathbf{r}}}{r^2} = 9 \times 10^9 \times \frac{(2 \times 10^{-10}) \cdot (0)}{2^2} = 0 \, \text{V} \) (since \( \cos \theta = 0 \), equatorial plane).

0 V
4.5 V
9 V
18 V
1

A parallel plate capacitor has plates of area \( 0.05 \, \text{m}^2 \) and separation 0.25 mm in air. What is its capacitance? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

\( C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 0.05}{0.25 \times 10^{-3}} = 1.77 \times 10^{-9} \, \text{F} = 1770 \, \text{pF} \).

1770 pF
1500 pF
2000 pF
1000 pF
1

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