Electrostatic Potential and Capacitance Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A spherical conductor of radius 15 cm has a charge of \( 6 \times 10^{-8} \, \text{C} \). What is the electric field at 40 cm from the center? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

For \( r = 0.4 \, \text{m} > R = 0.15 \, \text{m} \), \( E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2} \).

\( E = 9 \times 10^9 \times \frac{6 \times 10^{-8}}{(0.4)^2} = 9 \times 10^9 \times \frac{6 \times 10^{-8}}{0.16} = 3.375 \times 10^3 \, \text{N/C} \).

3 × 10³ N/C
3.5 × 10³ N/C
3.375 × 10³ N/C
4 × 10³ N/C
3

A \( 4 \, \mu\text{F} \) capacitor charged to \( 100 \, \text{V} \) is connected to an uncharged \( 4 \, \mu\text{F} \) capacitor. What is the final potential difference?

Initial charge: \( Q = 4 \times 10^{-6} \times 100 = 4 \times 10^{-4} \, \text{C} \).

Total capacitance: \( 4 + 4 = 8 \, \mu\text{F} \).

Final voltage: \( V = \frac{Q}{C} = \frac{4 \times 10^{-4}}{8 \times 10^{-6}} = 50 \, \text{V} \).

40 V
60 V
50 V
70 V
3

A positively charged particle moves from a point of higher potential to a point of lower potential in an electrostatic field. What is the sign of the work done by the electric field?

The electric field does work on a charge as \( W = q \Delta V \). For a positive charge moving from higher potential (\( V_h \)) to lower potential (\( V_l \)), \( \Delta V = V_l - V_h < 0 \). Since \( q > 0 \), the work done by the field \( W = q (V_l - V_h) \) is negative, indicating that the field opposes the motion or that the particle gains kinetic energy at the expense of potential energy, and the field does negative work if the motion is driven externally.

Zero
Positive
Negative
Depends on the path taken
3

A charge of \( 9 \, \mu\text{C} \) is moved from infinity to a point where the potential is \( 40 \, \text{V} \). What is the work done?

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

\( W = 9 \times 10^{-6} \times 40 = 3.6 \times 10^{-4} \, \text{J} = 0.36 \, \text{mJ} \).

0.36 mJ
0.4 mJ
0.3 mJ
0.5 mJ
1

In a system of two point charges of opposite signs placed close to each other, what can be said about the equipotential surface passing through the midpoint between them?

For two point charges of opposite signs (\( +q \) and \( -q \)), the potential at the midpoint is zero (since \( V = \frac{q}{4 \pi \varepsilon_0 r} - \frac{q}{4 \pi \varepsilon_0 r} = 0 \), where \( r \) is the distance from each charge to the midpoint). The equipotential surface where \( V = 0 \) is the plane perpendicular to the line joining the charges, passing through the midpoint, as points on this plane equidistant from both charges have potentials that cancel out.

It is a spherical surface centered at one charge
It is a curved surface bending toward the negative charge
It is a plane perpendicular to the line joining the charges
It is a cylindrical surface around the charges
3

A spherical conductor of radius 35 cm has a charge of \( 14 \times 10^{-8} \, \text{C} \). What is the electric field at 80 cm from the center? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

For \( r = 0.8 \, \text{m} > R = 0.35 \, \text{m} \), \( E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2} \).

\( E = 9 \times 10^9 \times \frac{14 \times 10^{-8}}{(0.8)^2} = 9 \times 10^9 \times \frac{14 \times 10^{-8}}{0.64} \approx 1.969 \times 10^3 \, \text{N/C} \).

1.8 × 10³ N/C
2 × 10³ N/C
1.969 × 10³ N/C
2.5 × 10³ N/C
3

A parallel plate capacitor with capacitance \( 250 \, \text{pF} \) has a dielectric (\( K = 4 \), thickness \( d/7 \)) inserted. What is the new capacitance? (Original separation \( d \)).

Potential difference: \( V = E_0 \left( \frac{6d}{7} \right) + \frac{E_0}{K} \left( \frac{d}{7} \right) = E_0 d \left( \frac{6}{7} + \frac{1}{7 \times 4} \right) \).

\( V = E_0 d \left( \frac{6}{7} + \frac{1}{28} \right) = E_0 d \times \frac{25}{28} \).

\( C = \frac{Q}{V} = \frac{Q}{\frac{25}{28} V_0} = \frac{28}{25} \times 250 = 280 \, \text{pF} \).

270 pF
275 pF
285 pF
280 pF
4

Two hollow spherical conductors of different radii are charged with the same amount of charge and then connected by a thin conducting wire. What happens to the charges on the conductors?

When two conductors are connected by a wire, they reach the same potential. The potential on the surface of a spherical conductor is given by \( V = \frac{Q}{4 \pi \varepsilon_0 R} \), where \( Q \) is the charge and \( R \) is the radius. If the radii are different, say \( R_1 \) and \( R_2 \), and initially both have charge \( Q \), after connection, \( V_1 = V_2 \), so \( \frac{Q_1}{R_1} = \frac{Q_2}{R_2} \). Since \( Q_1 + Q_2 = 2Q \), the charges redistribute such that the smaller sphere gets more charge due to its smaller radius (higher potential per unit charge), leading to a redistribution of charges.

Charges redistribute so that the potentials become equal
Charges remain the same on both conductors
All charge transfers to the larger conductor
All charge transfers to the smaller conductor
1

In a parallel combination of capacitors, why does each capacitor have the same potential difference across its plates?

In a parallel combination, the capacitors are connected across the same two nodes of the circuit. Since voltage (potential difference) is the same between these nodes (as they are directly connected to the same battery or voltage source), each capacitor experiences the same potential difference across its plates. The charge on each capacitor varies depending on its capacitance (\( Q = C V \)), but the voltage \( V \) remains the same for all capacitors in parallel.

Because they share the same charge
Because their capacitances are equal
Because they are connected across the same voltage source
Because the electric field between plates is identical
3

Two capacitors \( 10 \, \mu\text{F} \) and \( 20 \, \mu\text{F} \) are in parallel. What is the total capacitance?

\( C = C_1 + C_2 = 10 + 20 = 30 \, \mu\text{F} \).

25 µF
30 µF
35 µF
40 µF
2

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