Electric Charges and Fields Chapter-Wise Test 18

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A conducting sphere of radius 30 cm has a surface charge density of \( 50 \, \mu\text{C/m}^2 \). What is the total charge on it?

Surface area: \( A = 4 \pi r^2 = 4 \pi (0.3)^2 = 0.36 \pi \, \text{m}^2 \).

Charge: \( q = \sigma A = 50 \times 10^{-6} \times 0.36 \pi = 5.654 \times 10^{-5} \, \text{C} \).

\( 4.5 \times 10^{-5} \, \text{C} \)
\( 5.0 \times 10^{-5} \, \text{C} \)
\( 5.65 \times 10^{-5} \, \text{C} \)
\( 6.0 \times 10^{-5} \, \text{C} \)
3

What explains why the electric field due to a charged object can penetrate an insulator but not a conductor?

Insulators lack free charges to redistribute and cancel an external field, allowing it to penetrate. Conductors, with mobile charges, redistribute them to create an opposing field, shielding the interior from penetration.

Charge quantization
Field symmetry
Charge density
Charge mobility
4

A thin spherical shell of radius 12 cm has \( q = 10 \, \mu\text{C} \). What is the electric field at 14 cm from the center?

Outside shell: \( E = \frac{k q}{r^2} \).

\( E = 9 \times 10^9 \times \frac{10 \times 10^{-6}}{(0.14)^2} = 9 \times 10^9 \times \frac{10 \times 10^{-6}}{0.0196} = 4.59 \times 10^6 \, \text{N/C} \).

\( 4.2 \times 10^6 \, \text{N/C} \)
\( 4.4 \times 10^6 \, \text{N/C} \)
\( 4.5 \times 10^6 \, \text{N/C} \)
\( 4.59 \times 10^6 \, \text{N/C} \)
4

A thin spherical shell of radius 15 cm has \( q = 6 \, \mu\text{C} \). What is the electric field at 10 cm from the center?

Inside shell (\( r < R \)): \( E = 0 \) (Gauss’s law).

\( 1.5 \times 10^6 \, \text{N/C} \)
\( 1.0 \times 10^6 \, \text{N/C} \)
\( 0 \, \text{N/C} \)
\( 0.5 \times 10^6 \, \text{N/C} \)
3

What ensures that the electric field inside a hollow conductor remains zero even if an external charge is placed nearby?

Electrostatic shielding occurs as free charges in the conductor redistribute to cancel any external field inside. This adjustment creates an induced field that neutralizes the external influence, maintaining zero field within the hollow region.

Charge quantization
Electrostatic shielding
Field superposition
Charge symmetry
2

A dipole with \( p = 9 \times 10^{-9} \, \text{C m} \) is at 90° to a field \( E = 3 \times 10^4 \, \text{N/C} \). What is the torque?

\( \tau = p E \sin \theta \).

\( \tau = 9 \times 10^{-9} \times 3 \times 10^4 \times \sin 90^\circ = 27 \times 10^{-5} \times 1 = 2.7 \times 10^{-4} \, \text{N m} \).

\( 2.4 \times 10^{-4} \, \text{N m} \)
\( 2.5 \times 10^{-4} \, \text{N m} \)
\( 2.6 \times 10^{-4} \, \text{N m} \)
\( 2.7 \times 10^{-4} \, \text{N m} \)
4

What ensures that electric field lines never cross each other in a static field configuration?

The electric field has a unique direction at each point, defined by the force on a test charge. Crossing lines would imply multiple directions at one point, which is impossible in a static field where the field vector is well-defined.

Charge quantization
Field cancellation
Charge density
Unique direction
4

Two point charges \( 6 \times 10^{-7} \, \text{C} \) and \( 9 \times 10^{-7} \, \text{C} \) are 45 cm apart in air. What is the magnitude of the force between them?

Using Coulomb’s law: \( F = k \frac{|q_1 q_2|}{r^2} \).

\( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( q_1 = 6 \times 10^{-7} \, \text{C} \), \( q_2 = 9 \times 10^{-7} \, \text{C} \), \( r = 0.45 \, \text{m} \).

\( |q_1 q_2| = 6 \times 9 \times 10^{-14} = 54 \times 10^{-14} \, \text{C}^2 \).

\( r^2 = (0.45)^2 = 0.2025 \, \text{m}^2 \).

\( F = 9 \times 10^9 \times \frac{54 \times 10^{-14}}{0.2025} = 9 \times 10^9 \times 2.667 \times 10^{-12} = 0.024 \, \text{N} \).

0.024 N
0.027 N
0.030 N
0.033 N
1

Two point charges \( q_1 = 4 \, \mu\text{C} \) and \( q_2 = -6 \, \mu\text{C} \) are placed 20 cm apart in vacuum. What is the magnitude of the electrostatic force between them?

Using Coulomb's law: \( F = k \frac{|q_1 q_2|}{r^2} \).

Given: \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( q_1 = 4 \times 10^{-6} \, \text{C} \), \( q_2 = -6 \times 10^{-6} \, \text{C} \), \( r = 0.2 \, \text{m} \).

Magnitude: \( |q_1 q_2| = (4 \times 10^{-6}) \times (6 \times 10^{-6}) = 24 \times 10^{-12} \, \text{C}^2 \).

\( r^2 = (0.2)^2 = 0.04 \, \text{m}^2 \).

\( F = 9 \times 10^9 \times \frac{24 \times 10^{-12}}{0.04} = 9 \times 10^9 \times 6 \times 10^{-10} = 5.4 \, \text{N} \).

4.5 N
5.4 N
6.0 N
7.2 N
2

A closed surface has a net flux of \( 7.91 \times 10^5 \, \text{Nm}^2/\text{C} \). What is the charge enclosed?

\( \phi = \frac{q}{\varepsilon_0} \).

\( q = \phi \varepsilon_0 = 7.91 \times 10^5 \times 8.854 \times 10^{-12} = 7 \times 10^{-6} \, \text{C} = 7 \, \mu\text{C} \).

\( 6.8 \, \mu\text{C} \)
\( 6.9 \, \mu\text{C} \)
\( 7.0 \, \mu\text{C} \)
\( 7.1 \, \mu\text{C} \)
3

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