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  4. Electrostatic Potential and Capacitance Chapter-Wise Test 2

Electrostatic Potential and Capacitance Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

When two identical capacitors, one charged and one uncharged, are connected in parallel, why does the total energy decrease even though charge is conserved?

When a charged capacitor (\( C \), charge \( Q \), voltage \( V \)) is connected in parallel with an uncharged capacitor (\( C \)), the total capacitance becomes \( 2C \), and the charge redistributes to a final voltage \( V' = Q/(2C) = V/2 \). Initial energy is \( U_i = \frac{Q^2}{2C} \), while final energy is \( U_f = \frac{(Q)^2}{2 \times 2C} = \frac{Q^2}{4C} \), which is half the initial energy. The decrease occurs because energy is lost as heat or radiation during the transient current flow when charges redistribute, even though charge is conserved.

The capacitors mutually cancel their energies
Energy is lost as heat during charge redistribution
The total charge halves, reducing energy
The capacitors induce opposite potentials
2

A spherical conductor of radius 20 cm has a charge of \( 8 \times 10^{-8} \, \text{C} \). What is the electric field at 50 cm from the center? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

For \( r = 0.5 \, \text{m} > R = 0.2 \, \text{m} \), \( E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2} \).

\( E = 9 \times 10^9 \times \frac{8 \times 10^{-8}}{(0.5)^2} = 9 \times 10^9 \times \frac{8 \times 10^{-8}}{0.25} = 2.88 \times 10^3 \, \text{N/C} \).

2.5 × 10³ N/C
3 × 10³ N/C
2.88 × 10³ N/C
3.5 × 10³ N/C
3

Two capacitors of \( 15 \, \text{pF} \) each are connected in series. What is the equivalent capacitance?

\( \frac{1}{C} = \frac{1}{15} + \frac{1}{15} = \frac{2}{15} \).

\( C = \frac{15}{2} = 7.5 \, \text{pF} \).

5 pF
7.5 pF
10 pF
15 pF
2

A \( 5 \, \mu\text{F} \) capacitor is charged to \( 200 \, \text{V} \). What is the energy stored in it?

\( U = \frac{1}{2} C V^2 = \frac{1}{2} \times 5 \times 10^{-6} \times (200)^2 \).

\( U = \frac{1}{2} \times 5 \times 10^{-6} \times 40000 = 0.1 \, \text{J} \).

0.08 J
0.12 J
0.1 J
0.15 J
3

In an electrostatic field, if a positive charge is moved along an equipotential surface, what can be said about the work done by the electric field?

An equipotential surface has a constant potential at all points. The work done by the electric field when a charge moves along such a surface is zero because the potential difference between any two points on the surface is zero (\( W = q \Delta V \), and \( \Delta V = 0 \)). Additionally, the electric field is always perpendicular to an equipotential surface, so there is no component of the field along the direction of motion, confirming that no work is done.

It is zero
It depends on the path taken
It is proportional to the distance moved
It is negative
1

A point charge \( Q = 6 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 2 m away from the charge. (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Potential due to a point charge: \( V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r} \).

Substitute: \( V = 9 \times 10^9 \times \frac{6 \times 10^{-9}}{2} = 9 \times 10^9 \times 3 \times 10^{-9} = 27 \, \text{V} \).

27 V
30 V
24 V
20 V
1

A point charge \( Q = 21 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 7 m away from the charge. (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Potential due to a point charge: \( V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r} \).

Substitute: \( V = 9 \times 10^9 \times \frac{21 \times 10^{-9}}{7} = 9 \times 10^9 \times 3 \times 10^{-9} = 27 \, \text{V} \).

27 V
30 V
24 V
20 V
1

Why does the electric field due to an infinite charged sheet remain constant with distance from the sheet?

Using Gauss’s law for an infinite charged sheet with surface charge density \( \sigma \), a Gaussian surface (e.g., a cylinder perpendicular to the sheet) shows that the electric field \( E \) is perpendicular to the sheet and constant. The flux through the cylinder's end caps is \( E \cdot 2A \), and the enclosed charge is \( \sigma A \), so \( E \cdot 2A = \frac{\sigma A}{\varepsilon_0} \), giving \( E = \frac{\sigma}{2 \varepsilon_0} \). Since no distance term appears, \( E \) is independent of distance from the sheet.

The field depends only on surface charge density
The field decreases inversely with distance
The sheet's charge varies with distance
The field cancels out at large distances
1

When a conductor is placed in an external electric field, why does the potential throughout its volume become constant in electrostatic equilibrium?

In electrostatic equilibrium, the electric field inside a conductor is zero because free charges rearrange to cancel any internal field. Since the electric field is the negative gradient of potential (\( E = -\frac{dV}{dr} \)), if \( E = 0 \), the potential gradient must be zero, implying the potential \( V \) is constant throughout the conductor's volume. Any potential difference would drive charge movement, contradicting the equilibrium state.

Because the electric field inside is zero
Because charges accumulate only inside the conductor
Because the external field penetrates the conductor
Because the potential varies linearly inside
1

Three capacitors \( 2 \, \text{pF} \), \( 4 \, \text{pF} \), and \( 8 \, \text{pF} \) are in parallel. What is the total capacitance?

\( C = 2 + 4 + 8 = 14 \, \text{pF} \).

12 pF
16 pF
14 pF
18 pF
3

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