Electrostatic Potential and Capacitance Chapter-Wise Test 8

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Three charges \( +5 \, \mu\text{C} \), \( -2 \, \mu\text{C} \), and \( +3 \, \mu\text{C} \) are at \( (0, 0, 0) \), \( (5, 0, 0) \), and \( (0, 5, 0) \, \text{m} \). What is the potential at \( (5, 5, 0) \, \text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Distances: \( r_1 = \sqrt{5^2 + 5^2} = 5\sqrt{2} \, \text{m} \), \( r_2 = 5 \, \text{m} \), \( r_3 = 5 \, \text{m} \).

\( V = 9 \times 10^9 \left( \frac{5 \times 10^{-6}}{5\sqrt{2}} + \frac{-2 \times 10^{-6}}{5} + \frac{3 \times 10^{-6}}{5} \right) \).

\( V = 9 \times 10^9 \left( \frac{5 \times 10^{-6}}{7.07} - \frac{2 \times 10^{-6}}{5} + \frac{3 \times 10^{-6}}{5} \right) \).

\( V = 9 \times 10^9 \left( 0.707 \times 10^{-6} - 0.4 \times 10^{-6} + 0.6 \times 10^{-6} \right) \).

\( V = 9 \times 10^9 \times 0.907 \times 10^{-6} = 8163 \, \text{V} \).

8000 V
8163 V
8200 V
8500 V
2

Why does the potential energy of a system of two opposite charges increase when their separation decreases?

For two opposite charges \( +q \) and \( -q \), the potential energy is \( U = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r} = -\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r} \), negative due to attraction. Decreasing \( r \) makes \( U \) more negative (e.g., from \( -1/r_1 \) to \( -1/r_2 \), \( r_2 < r_1 \), \( U \) decreases). However, the question interprets "increase" as increasing in magnitude (more negative), as the attractive energy becomes stronger, lowering the system's energy state.

The charges repel more strongly
The field between them decreases
The potential increases with distance
The attractive interaction strengthens
4

A spherical conductor of radius 30 cm has a charge of \( 12 \times 10^{-8} \, \text{C} \). What is the electric field at 70 cm from the center? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

For \( r = 0.7 \, \text{m} > R = 0.3 \, \text{m} \), \( E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2} \).

\( E = 9 \times 10^9 \times \frac{12 \times 10^{-8}}{(0.7)^2} = 9 \times 10^9 \times \frac{12 \times 10^{-8}}{0.49} \approx 2.204 \times 10^3 \, \text{N/C} \).

2 × 10³ N/C
2.5 × 10³ N/C
2.204 × 10³ N/C
3 × 10³ N/C
3

A parallel plate capacitor has plates of area \( 0.02 \, \text{m}^2 \) separated by 2 mm in air. What is its capacitance? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

\( C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 0.02}{2 \times 10^{-3}} = 8.85 \times 10^{-11} \, \text{F} = 88.5 \, \text{pF} \).

80 pF
88.5 pF
90 pF
100 pF
2

Why does the capacitance of a parallel plate capacitor increase when a dielectric slab is inserted between the plates while keeping the plates connected to a battery?

When a dielectric slab (with \( K > 1 \)) is inserted while the capacitor remains connected to a battery (constant voltage \( V \)), the electric field decreases due to polarization (\( E = E_0/K \)), and the capacitance increases (\( C' = K C \)). This happens because the dielectric reduces the field, allowing more charge to be stored on the plates (\( Q = C V \), so \( Q' = K C V \)) for the same voltage, effectively increasing the capacitance as \( C = \frac{Q}{V} \).

The plates attract more charge due to increased voltage
The dielectric increases the plate area
The dielectric allows more charge storage for the same voltage
The dielectric reduces the plate separation
3

Two charges \( 10 \, \mu\text{C} \) and \( -5 \, \mu\text{C} \) are placed 20 cm apart. What is the potential energy of the system? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

\( U = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r} = 9 \times 10^9 \times \frac{10 \times 10^{-6} \times (-5 \times 10^{-6})}{0.2} = 9 \times 10^9 \times \frac{-50 \times 10^{-12}}{0.2} = -2.25 \, \text{J} \).

-2 J
-2.5 J
-2.25 J
-3 J
3

Two charges \( 8 \, \mu\text{C} \) and \( -4 \, \mu\text{C} \) are at \( (6, 0, 0) \) and \( (-6, 0, 0) \, \text{cm} \). 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 = 0.06 m.

\( V = 9 \times 10^9 \left( \frac{8 \times 10^{-6}}{0.06} + \frac{-4 \times 10^{-6}}{0.06} \right) = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{0.06} \).

\( V = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{0.06} = 6 \times 10^5 \, \text{V} \).

5 × 10⁵ V
5.5 × 10⁵ V
6.5 × 10⁵ V
6 × 10⁵ V
4

A parallel plate capacitor with \( C = 100 \, \text{pF} \) in air has a dielectric (\( K = 10 \)) inserted fully between plates. What is the new capacitance?

\( C' = K C = 10 \times 100 = 1000 \, \text{pF} \).

900 pF
1000 pF
1100 pF
1200 pF
2

A metal sphere is placed inside a uniform electric field \( E \). What is the electric field inside the sphere after it reaches electrostatic equilibrium?

In electrostatic equilibrium, the electric field inside a conductor is zero. When a metal sphere is placed in a uniform electric field, charges redistribute on its surface such that the induced field inside cancels the external field. This results in a net electric field of zero inside the sphere, regardless of the external field strength, as conductors allow free movement of charges to achieve this equilibrium.

It depends on the strength of the external field
It is zero due to charge redistribution
It is equal to the external field
It is non-uniform and depends on the sphere's radius
2

A point charge \( Q = 8 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 4 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: \( V = 9 \times 10^9 \times \frac{8 \times 10^{-9}}{4} = 9 \times 10^9 \times 2 \times 10^{-9} = 18 \, \text{V} \).

18 V
15 V
20 V
22 V
1

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