Electric Charges and Fields Chapter-Wise Test 13

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What explains why the electric field outside a neutral object can be non-zero when placed in an external field?

Induced charge separation occurs in a neutral object under an external field, creating dipoles or surface charges. These induced charges produce their own field, which adds to the external field, resulting in a non-zero total field outside.

Charge quantization
Field cancellation
Charge symmetry
Induced charge separation
4

A uniform electric field \( E = 5 \times 10^3 \, \text{N/C} \) is along the y-axis. What is the flux through a square of side 40 cm in the xz-plane?

Area vector \( \Delta \mathbf{S} = (0.4)^2 = 0.16 \, \text{m}^2 \) along y-axis.

Flux: \( \phi = \mathbf{E} \cdot \Delta \mathbf{S} = 5 \times 10^3 \times 0.16 = 800 \, \text{Nm}^2/\text{C} \).

\( 800 \, \text{Nm}^2/\text{C} \)
\( 850 \, \text{Nm}^2/\text{C} \)
\( 900 \, \text{Nm}^2/\text{C} \)
\( 950 \, \text{Nm}^2/\text{C} \)
1

What fundamental principle allows the electric field due to multiple charges to be calculated as the vector sum of the fields due to each charge individually?

The superposition principle states that the electric field produced by multiple charges is the vector sum of the fields produced by each charge independently. This principle holds because electric forces and fields are linear and unaffected by the presence of other charges.

Conservation of charge
Superposition
Coulomb’s law
Gauss’s law
2

A conducting sphere of radius 19 cm has an electric field of \( 3 \times 10^3 \, \text{N/C} \) at 38 cm from its center. What is the charge?

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

\( 3 \times 10^3 = 9 \times 10^9 \times \frac{q}{(0.38)^2} \).

\( q = \frac{3 \times 10^3 \times 0.1444}{9 \times 10^9} = 4.81 \times 10^{-8} \, \text{C} \).

\( 4.5 \times 10^{-8} \, \text{C} \)
\( 4.7 \times 10^{-8} \, \text{C} \)
\( 4.8 \times 10^{-8} \, \text{C} \)
\( 4.81 \times 10^{-8} \, \text{C} \)
4

Two point charges \( 9 \times 10^{-7} \, \text{C} \) and \( -3 \times 10^{-7} \, \text{C} \) are 100 cm apart in vacuum. What is the magnitude of the force between them?

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

\( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( q_1 = 9 \times 10^{-7} \, \text{C} \), \( q_2 = -3 \times 10^{-7} \, \text{C} \), \( r = 1.0 \, \text{m} \).

\( |q_1 q_2| = 9 \times 3 \times 10^{-14} = 27 \times 10^{-14} \, \text{C}^2 \).

\( r^2 = (1.0)^2 = 1 \, \text{m}^2 \).

\( F = 9 \times 10^9 \times \frac{27 \times 10^{-14}}{1} = 9 \times 10^9 \times 2.7 \times 10^{-13} = 0.00243 \, \text{N} \).

0.00243 N
0.0025 N
0.0027 N
0.0030 N
1

A thin spherical shell of radius 25 cm has a charge of \( 15 \, \mu\text{C} \). What is the electric field at a point 30 cm from the center?

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

\( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( q = 15 \times 10^{-6} \, \text{C} \), \( r = 0.3 \, \text{m} \).

\( E = 9 \times 10^9 \times \frac{15 \times 10^{-6}}{(0.3)^2} = 9 \times 10^9 \times \frac{15 \times 10^{-6}}{0.09} = 1.5 \times 10^6 \, \text{N/C} \).

\( 1.5 \times 10^6 \, \text{N/C} \)
\( 1.8 \times 10^6 \, \text{N/C} \)
\( 2.0 \times 10^6 \, \text{N/C} \)
\( 2.2 \times 10^6 \, \text{N/C} \)
1

A dipole with \( p = 4 \times 10^{-9} \, \text{C m} \) is at 30° to a field \( E = 6 \times 10^4 \, \text{N/C} \). What is the torque?

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

\( \tau = 4 \times 10^{-9} \times 6 \times 10^4 \times \sin 30^\circ = 24 \times 10^{-5} \times 0.5 = 1.2 \times 10^{-4} \, \text{N m} \).

\( 1.0 \times 10^{-4} \, \text{N m} \)
\( 1.1 \times 10^{-4} \, \text{N m} \)
\( 1.15 \times 10^{-4} \, \text{N m} \)
\( 1.2 \times 10^{-4} \, \text{N m} \)
4

A thin spherical shell of radius 14 cm has a charge of \( 7 \, \mu\text{C} \). What is the electric field at a point 16 cm from the center?

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

\( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( q = 7 \times 10^{-6} \, \text{C} \), \( r = 0.16 \, \text{m} \).

\( E = 9 \times 10^9 \times \frac{7 \times 10^{-6}}{(0.16)^2} = 9 \times 10^9 \times \frac{7 \times 10^{-6}}{0.0256} = 2.46 \times 10^6 \, \text{N/C} \).

\( 2.46 \times 10^6 \, \text{N/C} \)
\( 2.5 \times 10^6 \, \text{N/C} \)
\( 2.7 \times 10^6 \, \text{N/C} \)
\( 3.0 \times 10^6 \, \text{N/C} \)
1

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

For an infinite plane sheet, symmetry and Gauss’s law show that the field depends only on surface charge density, not distance. The field lines are parallel and uniform, as contributions from all parts of the infinite sheet balance out distance effects.

Charge cancellation
Field superposition
Infinite extent
Charge mobility
3

A thin spherical shell of radius 17 cm has \( q = 13 \, \mu\text{C} \). What is the electric field at 20 cm from the center?

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

\( E = 9 \times 10^9 \times \frac{13 \times 10^{-6}}{(0.2)^2} = 9 \times 10^9 \times \frac{13 \times 10^{-6}}{0.04} = 2.925 \times 10^6 \, \text{N/C} \).

\( 2.8 \times 10^6 \, \text{N/C} \)
\( 2.85 \times 10^6 \, \text{N/C} \)
\( 2.9 \times 10^6 \, \text{N/C} \)
\( 2.925 \times 10^6 \, \text{N/C} \)
4

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