Electrostatic Potential and Capacitance Chapter-Wise Test 10

Correct answer Carries: 4.

Wrong Answer Carries: -1.

An electric dipole with moment \( p = 6 \times 10^{-9} \, \text{C m} \) lies along the z-axis. What is the potential at \( (0, 0, -2) \, \text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

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

\( V = -9 \times 10^9 \times \frac{6 \times 10^{-9}}{2^2} = -9 \times 10^9 \times \frac{6 \times 10^{-9}}{4} = -13.5 \, \text{V} \).

-13.5 V
13.5 V
0 V
-10 V
1

In a system of two capacitors connected in series across a battery, why does the capacitor with smaller capacitance store more energy compared to the one with larger capacitance?

In series, the charge \( Q \) on each capacitor is the same. Energy stored in a capacitor is \( U = \frac{Q^2}{2C} \). For a smaller capacitance \( C \), the denominator \( 2C \) is smaller, so \( U \) is larger compared to a capacitor with larger \( C \). Alternatively, since \( V = \frac{Q}{C} \), the smaller capacitor has a larger voltage \( V \), and energy \( U = \frac{1}{2} C V^2 \) increases with \( V^2 \), leading to more energy in the smaller capacitor.

It has a larger voltage across it
It stores a larger charge
Its electric field is weaker
Its plates are closer together
1

Why does the electric field inside a uniformly charged spherical shell remain zero even if the shell is placed in an external uniform electric field?

For a uniformly charged spherical shell in electrostatic equilibrium, the electric field inside is zero regardless of external fields due to electrostatic shielding. Applying Gauss’s law inside the shell (no charge enclosed within the cavity), \( E = 0 \). The external field induces charge redistribution on the shell's outer surface, but this does not affect the interior, as the induced charges ensure the internal field cancels out, maintaining \( E = 0 \) inside the cavity.

The external field penetrates the shell
Charge redistribution cancels the internal field
The shell's charge neutralizes the field
The field lines bend around the shell
2

A spherical conductor of radius 5 cm has a charge of \( 5 \times 10^{-8} \, \text{C} \). 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: \( V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R} \).

\( V = 9 \times 10^9 \times \frac{5 \times 10^{-8}}{0.05} = 9 \times 10^9 \times 10^{-6} = 9000 \, \text{V} \).

9000 V
8000 V
10000 V
12000 V
1

In a series combination of capacitors with different dielectric materials between their plates, why do capacitors with higher dielectric constants have lower potential differences?

In series, the charge \( Q \) on each capacitor is the same. Capacitance is \( C = \frac{K \varepsilon_0 A}{d} \), where \( K \) is the dielectric constant. A higher \( K \) increases \( C \). Since \( V = \frac{Q}{C} \), a larger \( C \) (due to higher \( K \)) results in a smaller \( V \). Thus, capacitors with higher dielectric constants have lower potential differences because their capacitance is larger for the same \( Q \), distributing the voltage inversely with capacitance.

They store more charge
Their plates are closer together
They have lower capacitance
They have higher capacitance
4

Why does the total energy stored in a system of capacitors connected in series decrease if one of the capacitors is removed and the battery remains connected?

In series, the equivalent capacitance \( C_{\text{eq}} \) decreases when a capacitor is removed (\( \frac{1}{C_{\text{eq}}} = \sum \frac{1}{C_i} \)). With a battery maintaining constant voltage \( V \), the energy stored is \( U = \frac{1}{2} C_{\text{eq}} V^2 \). A smaller \( C_{\text{eq}} \) reduces \( U \), as energy is directly proportional to capacitance under constant \( V \). Additionally, removing a capacitor redistributes charges, potentially dissipating energy during the process, further reducing total energy.

Equivalent capacitance decreases
Voltage across the system increases
Charge on remaining capacitors decreases
Electric field between plates increases
1

A spherical conductor of radius 10 cm has a charge of \( 5 \times 10^{-8} \, \text{C} \). What is the electric field at 25 cm from the center? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

For \( r = 0.25 \, \text{m} > R = 0.1 \, \text{m} \), \( E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2} \).

\( E = 9 \times 10^9 \times \frac{5 \times 10^{-8}}{(0.25)^2} = 9 \times 10^9 \times \frac{5 \times 10^{-8}}{0.0625} = 7.2 \times 10^3 \, \text{N/C} \).

6 × 10³ N/C
8 × 10³ N/C
7.2 × 10³ N/C
9 × 10³ N/C
3

Three charges \( +7 \, \mu\text{C} \), \( -4 \, \mu\text{C} \), and \( +2 \, \mu\text{C} \) are at \( (0, 0, 0) \), \( (7, 0, 0) \), and \( (0, 7, 0) \, \text{m} \). What is the potential at \( (7, 7, 0) \, \text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Distances: \( r_1 = \sqrt{7^2 + 7^2} = 7\sqrt{2} \, \text{m} \), \( r_2 = 7 \, \text{m} \), \( r_3 = 7 \, \text{m} \).

\( V = 9 \times 10^9 \left( \frac{7 \times 10^{-6}}{7\sqrt{2}} + \frac{-4 \times 10^{-6}}{7} + \frac{2 \times 10^{-6}}{7} \right) \).

\( V = 9 \times 10^9 \left( \frac{7 \times 10^{-6}}{9.899} - \frac{4 \times 10^{-6}}{7} + \frac{2 \times 10^{-6}}{7} \right) \).

\( V = 9 \times 10^9 \left( 0.707 \times 10^{-6} - 0.571 \times 10^{-6} + 0.286 \times 10^{-6} \right) \).

\( V = 9 \times 10^9 \times 0.422 \times 10^{-6} = 3798 \, \text{V} \).

3600 V
3798 V
3800 V
4000 V
2

A charged particle is released from rest near a large charged conducting sheet. Why does the particle initially accelerate toward or away from the sheet?

A large charged conducting sheet creates a uniform electric field near its surface (\( E = \frac{\sigma}{\varepsilon_0} \)). If the sheet is positively charged, the field points away from it; if negatively charged, it points toward it. A charged particle experiences a force \( F = q E \). Depending on the particle's charge (same or opposite to the sheet), it accelerates toward (opposite charges) or away (like charges) due to this force, as the field exerts an attractive or repulsive force.

Due to gravitational attraction
Due to induced charges on the particle
Due to varying potential gradient
Due to the uniform electric field of the sheet
4

An electric dipole with moment \( p = 5 \times 10^{-9} \, \text{C m} \) lies along the y-axis. What is the potential at \( (0, 3, 0) \, \text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Along the dipole axis (\( \theta = 0^\circ \)): \( V = \frac{1}{4 \pi \varepsilon_0} \frac{p}{r^2} \).

\( V = 9 \times 10^9 \times \frac{5 \times 10^{-9}}{3^2} = 9 \times 10^9 \times \frac{5 \times 10^{-9}}{9} = 5 \, \text{V} \).

5 V
4 V
6 V
3 V
1

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