Electrostatic Potential and Capacitance Chapter-Wise Test 6

Correct answer Carries: 4.

Wrong Answer Carries: -1.

In a system of two parallel plate capacitors with different dielectrics connected in parallel, why does the capacitor with a higher dielectric constant store more energy for the same applied voltage?

In parallel, both capacitors have the same voltage \( V \). Capacitance is \( C = \frac{K \varepsilon_0 A}{d} \), where \( K \) is the dielectric constant. A higher \( K \) increases \( C \). Energy stored is \( U = \frac{1}{2} C V^2 \), so a larger \( C \) (due to higher \( K \)) leads to more energy stored for the same \( V \). The higher \( K \) allows more charge (\( Q = C V \)) to be stored, increasing energy without changing \( V \).

It has a smaller capacitance
It stores more energy due to larger capacitance
The electric field is stronger
The voltage across it is higher
2

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

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

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

-9 V
9 V
0 V
-6 V
1

What is the work done in moving a \( 5 \, \mu\text{C} \) charge from infinity to a point where the potential is \( 2000 \, \text{V} \)?

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

\( W = 5 \times 10^{-6} \times 2000 = 10^{-2} \, \text{J} = 0.01 \, \text{J} \).

0.005 J
0.01 J
0.015 J
0.02 J
2

A \( 12 \, \mu\text{F} \) capacitor charged to \( 30 \, \text{V} \) is connected to an uncharged \( 12 \, \mu\text{F} \) capacitor. What is the final potential difference?

Initial charge: \( Q = 12 \times 10^{-6} \times 30 = 3.6 \times 10^{-4} \, \text{C} \).

Total capacitance: \( 12 + 12 = 24 \, \mu\text{F} \).

Final voltage: \( V = \frac{Q}{C} = \frac{3.6 \times 10^{-4}}{24 \times 10^{-6}} = 15 \, \text{V} \).

10 V
20 V
15 V
25 V
3

Three charges \( +6 \, \mu\text{C} \), \( -3 \, \mu\text{C} \), and \( +1 \, \mu\text{C} \) are at \( (0, 0, 0) \), \( (6, 0, 0) \), and \( (0, 6, 0) \, \text{m} \). What is the potential at \( (6, 6, 0) \, \text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Distances: \( r_1 = \sqrt{6^2 + 6^2} = 6\sqrt{2} \, \text{m} \), \( r_2 = 6 \, \text{m} \), \( r_3 = 6 \, \text{m} \).

\( V = 9 \times 10^9 \left( \frac{6 \times 10^{-6}}{6\sqrt{2}} + \frac{-3 \times 10^{-6}}{6} + \frac{1 \times 10^{-6}}{6} \right) \).

\( V = 9 \times 10^9 \left( \frac{6 \times 10^{-6}}{8.485} - \frac{3 \times 10^{-6}}{6} + \frac{1 \times 10^{-6}}{6} \right) \).

\( V = 9 \times 10^9 \left( 0.707 \times 10^{-6} - 0.5 \times 10^{-6} + 0.167 \times 10^{-6} \right) \).

\( V = 9 \times 10^9 \times 0.374 \times 10^{-6} = 3366 \, \text{V} \).

3200 V
3366 V
3400 V
3500 V
2

A conductor has a surface charge density of \( 2.5 \times 10^{-6} \, \text{C/m}^2 \). What is the electric field just outside it? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

\( E = \frac{\sigma}{\varepsilon_0} = \frac{2.5 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 2.82 \times 10^5 \, \text{N/C} \).

2.5 × 10⁵ N/C
2.82 × 10⁵ N/C
3 × 10⁵ N/C
3.5 × 10⁵ N/C
2

A conductor has a surface charge density of \( 5 \times 10^{-7} \, \text{C/m}^2 \). What is the electric field just outside it? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

\( E = \frac{\sigma}{\varepsilon_0} = \frac{5 \times 10^{-7}}{8.85 \times 10^{-12}} \approx 5.65 \times 10^4 \, \text{N/C} \).

5 × 10⁴ N/C
5.65 × 10⁴ N/C
6 × 10⁴ N/C
7 × 10⁴ N/C
2

Two charges \( 24 \, \mu\text{C} \) and \( -12 \, \mu\text{C} \) are placed 24 cm apart. What is the potential energy of the system? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

\( U = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r} = 9 \times 10^9 \times \frac{24 \times 10^{-6} \times (-12 \times 10^{-6})}{0.24} \).

\( U = 9 \times 10^9 \times \frac{-288 \times 10^{-12}}{0.24} = -10.8 \, \text{J} \).

-10 J
-11 J
-10.8 J
-12 J
3

A conductor has a surface charge density of \( 3 \times 10^{-6} \, \text{C/m}^2 \). What is the electric field just outside it? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

\( E = \frac{\sigma}{\varepsilon_0} = \frac{3 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 3.39 \times 10^5 \, \text{N/C} \).

3 × 10⁵ N/C
3.39 × 10⁵ N/C
4 × 10⁵ N/C
5 × 10⁵ N/C
2

A parallel plate capacitor with capacitance \( 100 \, \text{pF} \) has a dielectric (\( K = 3 \), thickness \( d/3 \)) inserted. What is the new capacitance? (Original separation \( d \)).

Potential difference: \( V = E_0 \left( \frac{2d}{3} \right) + \frac{E_0}{K} \left( \frac{d}{3} \right) = E_0 d \left( \frac{2}{3} + \frac{1}{3 \times 3} \right) \).

\( V = E_0 d \left( \frac{2}{3} + \frac{1}{9} \right) = E_0 d \times \frac{7}{9} \).

\( C = \frac{Q}{V} = \frac{Q}{\frac{7}{9} V_0} = \frac{9}{7} \times 100 \approx 128.57 \, \text{pF} \).

120 pF
125 pF
130 pF
128.57 pF
4

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