Electric Charges and Fields Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What ensures that the electric field is always radial for an infinitely long, uniformly charged wire?

The cylindrical symmetry of an infinitely long, uniformly charged wire means the field must be radial (perpendicular to the wire) and uniform along its length. Any non-radial component would violate symmetry, and Gauss’s law confirms this radial dependence.

Charge quantization
Field gradient
Symmetry
Superposition
3

What principle explains why the net electric flux through a closed surface depends only on the charge enclosed within it?

Gauss’s law states that the electric flux through a closed surface is proportional to the net charge enclosed, regardless of its position or distribution inside. External charges do not contribute to the net flux due to the inverse-square nature of the field.

Coulomb’s law
Gauss’s law
Superposition principle
Charge conservation
2

An infinite line charge produces \( E = 3.6 \times 10^5 \, \text{N/C} \) at 5 cm. What is \( \lambda \)?

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

\( 3.6 \times 10^5 = \frac{2 \times 9 \times 10^9 \times \lambda}{0.05} \).

\( \lambda = \frac{3.6 \times 10^5 \times 0.05}{18 \times 10^9} = 1 \times 10^{-6} \, \text{C/m} \).

\( 0.8 \times 10^{-6} \, \text{C/m} \)
\( 0.9 \times 10^{-6} \, \text{C/m} \)
\( 0.95 \times 10^{-6} \, \text{C/m} \)
\( 1.0 \times 10^{-6} \, \text{C/m} \)
4

Two charges \( +7 \, \mu\text{C} \) and \( -5 \, \mu\text{C} \) are 60 cm apart. What is the electric field magnitude at the midpoint?

Midpoint distance = 30 cm = 0.3 m.

\( E_1 = 9 \times 10^9 \times \frac{7 \times 10^{-6}}{(0.3)^2} = 7 \times 10^5 \, \text{N/C} \) (towards \( -5 \, \mu\text{C} \)).

\( E_2 = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{(0.3)^2} = 5 \times 10^5 \, \text{N/C} \) (towards \( -5 \, \mu\text{C} \)).

Net \( E = 7 \times 10^5 + 5 \times 10^5 = 1.2 \times 10^6 \, \text{N/C} \).

\( 1.1 \times 10^6 \, \text{N/C} \)
\( 1.2 \times 10^6 \, \text{N/C} \)
\( 1.3 \times 10^6 \, \text{N/C} \)
\( 1.4 \times 10^6 \, \text{N/C} \)
2

An infinite line charge has \( E = 4.5 \times 10^5 \, \text{N/C} \) at 4 cm. What is \( \lambda \)?

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

\( 4.5 \times 10^5 = \frac{2 \times 9 \times 10^9 \times \lambda}{0.04} \).

\( \lambda = \frac{4.5 \times 10^5 \times 0.04}{18 \times 10^9} = 1 \times 10^{-6} \, \text{C/m} \).

\( 0.8 \times 10^{-6} \, \text{C/m} \)
\( 0.9 \times 10^{-6} \, \text{C/m} \)
\( 0.95 \times 10^{-6} \, \text{C/m} \)
\( 1.0 \times 10^{-6} \, \text{C/m} \)
4

A uniform field \( E = 7 \times 10^3 \, \text{N/C} \) is along the z-axis. What is the flux through a circle of radius 10 cm in the xy-plane?

Area: \( A = \pi (0.1)^2 = 0.0314 \, \text{m}^2 \).

Flux: \( \phi = E A \cos 0^\circ = 7 \times 10^3 \times 0.0314 = 219.8 \, \text{Nm}^2/\text{C} \).

200 \( \text{Nm}^2/\text{C} \)
210 \( \text{Nm}^2/\text{C} \)
220 \( \text{Nm}^2/\text{C} \)
230 \( \text{Nm}^2/\text{C} \)
3

A plane sheet has \( \sigma = 8.85 \times 10^{-11} \, \text{C/m}^2 \). What is the electric field near it?

\( E = \frac{\sigma}{2 \varepsilon_0} \).

\( E = \frac{8.85 \times 10^{-11}}{2 \times 8.854 \times 10^{-12}} = 5 \, \text{N/C} \).

\( 4.5 \, \text{N/C} \)
\( 5.0 \, \text{N/C} \)
\( 5.5 \, \text{N/C} \)
\( 6.0 \, \text{N/C} \)
2

A thin spherical shell of radius 32 cm has \( q = 9 \, \mu\text{C} \). What is the electric field at 25 cm from the center?

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

\( 1.5 \times 10^6 \, \text{N/C} \)
\( 1.2 \times 10^6 \, \text{N/C} \)
\( 0 \, \text{N/C} \)
\( 0.8 \times 10^6 \, \text{N/C} \)
3

A point charge \( 7 \, \mu\text{C} \) is at the origin. What is the electric field magnitude at a point 4 m along the y-axis?

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

\( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( q = 7 \times 10^{-6} \, \text{C} \), \( r = 4 \, \text{m} \).

\( E = 9 \times 10^9 \times \frac{7 \times 10^{-6}}{(4)^2} = 9 \times 10^9 \times \frac{7 \times 10^{-6}}{16} = 3.9375 \times 10^3 \, \text{N/C} \).

\( 3.94 \times 10^3 \, \text{N/C} \)
\( 4.5 \times 10^3 \, \text{N/C} \)
\( 5.0 \times 10^3 \, \text{N/C} \)
\( 5.6 \times 10^3 \, \text{N/C} \)
1

Three charges \( +6 \, \mu\text{C}, -3 \, \mu\text{C}, +3 \, \mu\text{C} \) are at the vertices of an equilateral triangle of side 1 m. What is the force magnitude on \( +6 \, \mu\text{C} \)?

\( F_1 = 9 \times 10^9 \times \frac{6 \times 3 \times 10^{-12}}{1^2} = 0.162 \, \text{N} \) (attractive).

\( F_2 = 9 \times 10^9 \times \frac{6 \times 3 \times 10^{-12}}{1^2} = 0.162 \, \text{N} \) (repulsive).

Angle 60°. Net \( F = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos 60^\circ} = \sqrt{0.162^2 + 0.162^2 + 0.026244} = 0.28 \, \text{N} \).

0.25 N
0.27 N
0.28 N
0.30 N
3

What ensures that the electric field outside a uniformly charged spherical shell behaves as if all charge were concentrated at its center?

Spherical symmetry and Gauss’s law show that for a uniform spherical shell, the field outside depends only on the total charge and distance, mimicking a point charge at the center. Inside, symmetry cancels the field, leaving zero.

Charge quantization
Field cancellation
Charge distribution
Spherical symmetry
4

A body loses \( 8.0 \times 10^{-8} \, \text{C} \) of charge when rubbed. How many electrons were transferred from it?

Losing charge means electrons are removed, so charge is positive.

\( q = n e \), \( e = 1.6 \times 10^{-19} \, \text{C} \).

\( n = \frac{q}{|e|} = \frac{8.0 \times 10^{-8}}{1.6 \times 10^{-19}} = 5 \times 10^{11} \).

\( 5 \times 10^{11} \)
\( 5.5 \times 10^{11} \)
\( 6 \times 10^{11} \)
\( 6.5 \times 10^{11} \)
1

A thin spherical shell of radius 18 cm has a charge of \( 12 \, \mu\text{C} \). What is the electric field at a point 20 cm from the center?

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

\( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( q = 12 \times 10^{-6} \, \text{C} \), \( r = 0.2 \, \text{m} \).

\( E = 9 \times 10^9 \times \frac{12 \times 10^{-6}}{(0.2)^2} = 9 \times 10^9 \times \frac{12 \times 10^{-6}}{0.04} = 2.7 \times 10^6 \, \text{N/C} \).

\( 2.7 \times 10^6 \, \text{N/C} \)
\( 3.0 \times 10^6 \, \text{N/C} \)
\( 3.3 \times 10^6 \, \text{N/C} \)
\( 3.6 \times 10^6 \, \text{N/C} \)
1

A point charge \( 15 \, \mu\text{C} \) is at the origin. What is the electric field magnitude at a point 6 m along the y-axis?

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

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

\( E = 9 \times 10^9 \times \frac{15 \times 10^{-6}}{(6)^2} = 9 \times 10^9 \times \frac{15 \times 10^{-6}}{36} = 3.75 \times 10^3 \, \text{N/C} \).

\( 3.75 \times 10^3 \, \text{N/C} \)
\( 4.0 \times 10^3 \, \text{N/C} \)
\( 4.2 \times 10^3 \, \text{N/C} \)
\( 4.5 \times 10^3 \, \text{N/C} \)
1

A body loses \( 3.2 \times 10^{-8} \, \text{C} \) of charge when rubbed. How many electrons were transferred from it?

Losing charge means electrons are removed, so charge is positive.

\( q = n e \), \( e = 1.6 \times 10^{-19} \, \text{C} \).

\( n = \frac{q}{|e|} = \frac{3.2 \times 10^{-8}}{1.6 \times 10^{-19}} = 2 \times 10^{11} \).

\( 2 \times 10^{11} \)
\( 2.5 \times 10^{11} \)
\( 3 \times 10^{11} \)
\( 4 \times 10^{11} \)
1

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