Electric Charges and Fields Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A closed surface has a net flux of \( 6.78 \times 10^5 \, \text{Nm}^2/\text{C} \). What is the charge enclosed?

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

\( q = \phi \varepsilon_0 = 6.78 \times 10^5 \times 8.854 \times 10^{-12} = 6 \times 10^{-6} \, \text{C} = 6 \, \mu\text{C} \).

\( 5.8 \, \mu\text{C} \)
\( 5.9 \, \mu\text{C} \)
\( 6.0 \, \mu\text{C} \)
\( 6.1 \, \mu\text{C} \)
3

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 25 cm in the xy-plane?

Area: \( A = \pi (0.25)^2 = 0.1963 \, \text{m}^2 \).

Flux: \( \phi = E A \cos 0^\circ = 7 \times 10^3 \times 0.1963 = 1374 \, \text{Nm}^2/\text{C} \).

1300 \( \text{Nm}^2/\text{C} \)
1350 \( \text{Nm}^2/\text{C} \)
1374 \( \text{Nm}^2/\text{C} \)
1400 \( \text{Nm}^2/\text{C} \)
3

A charge of \( 11 \, \mu\text{C} \) is at the center of a cube of edge 45 cm. What is the total flux through the cube?

Total flux: \( \phi = \frac{q}{\varepsilon_0} \).

\( \phi = \frac{11 \times 10^{-6}}{8.854 \times 10^{-12}} = 1.242 \times 10^6 \, \text{Nm}^2/\text{C} \).

\( 1.1 \times 10^6 \, \text{Nm}^2/\text{C} \)
\( 1.242 \times 10^6 \, \text{Nm}^2/\text{C} \)
\( 1.3 \times 10^6 \, \text{Nm}^2/\text{C} \)
\( 1.4 \times 10^6 \, \text{Nm}^2/\text{C} \)
2

A glass rod loses \( 6.4 \times 10^{-8} \, \text{C} \) of charge when rubbed with silk. How many electrons were transferred from it?

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

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

\( n = \frac{q}{|e|} = \frac{6.4 \times 10^{-8}}{1.6 \times 10^{-19}} = 4 \times 10^{11} \).

\( 4 \times 10^{11} \)
\( 4.5 \times 10^{11} \)
\( 5 \times 10^{11} \)
\( 5.5 \times 10^{11} \)
1

Three charges \( +8 \, \mu\text{C}, -4 \, \mu\text{C}, +2 \, \mu\text{C} \) are at the vertices of an equilateral triangle of side 1.8 m. What is the force magnitude on \( +8 \, \mu\text{C} \)?

\( F_1 = 9 \times 10^9 \times \frac{8 \times 4 \times 10^{-12}}{(1.8)^2} = 0.0889 \, \text{N} \) (attractive).

\( F_2 = 9 \times 10^9 \times \frac{8 \times 2 \times 10^{-12}}{(1.8)^2} = 0.0444 \, \text{N} \) (repulsive).

Angle 60°. Net \( F = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos 60^\circ} = \sqrt{0.0889^2 + 0.0444^2 + 0.00395} = 0.108 \, \text{N} \).

0.09 N
0.10 N
0.108 N
0.11 N
3

Which principle underlies the fact that the electric field due to a system of charges can be analyzed by considering each charge independently?

The superposition principle allows the total electric field to be the vector sum of fields from individual charges, assuming each field is unaffected by others. This linearity stems from the nature of electrostatic forces, enabling independent analysis.

Superposition
Charge quantization
Conservation of charge
Field symmetry
1

Two charges \( +5 \, \mu\text{C} \) and \( -7 \, \mu\text{C} \) are 35 cm apart. What is the electric field magnitude at the midpoint?

Midpoint distance = 17.5 cm = 0.175 m.

\( E_1 = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{(0.175)^2} = 1.47 \times 10^6 \, \text{N/C} \) (towards \( -7 \, \mu\text{C} \)).

\( E_2 = 9 \times 10^9 \times \frac{7 \times 10^{-6}}{(0.175)^2} = 2.06 \times 10^6 \, \text{N/C} \) (towards \( -7 \, \mu\text{C} \)).

Net \( E = 1.47 \times 10^6 + 2.06 \times 10^6 = 3.53 \times 10^6 \, \text{N/C} \).

\( 3.0 \times 10^6 \, \text{N/C} \)
\( 3.53 \times 10^6 \, \text{N/C} \)
\( 3.8 \times 10^6 \, \text{N/C} \)
\( 4.0 \times 10^6 \, \text{N/C} \)
2

Why does the electric field due to a uniformly charged infinite plane remain unaffected by the distance from the plane?

The infinite extent of the plane ensures uniform field distribution, as contributions from all parts balance out distance effects. Gauss’s law confirms the field is \( \sigma/2\varepsilon_0 \), independent of distance due to symmetry.

Charge quantization
Infinite extent
Field cancellation
Charge mobility
2

A thin spherical shell of radius 22 cm has \( q = 15 \, \mu\text{C} \). What is the electric field at 26 cm from the center?

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

\( E = 9 \times 10^9 \times \frac{15 \times 10^{-6}}{(0.26)^2} = 9 \times 10^9 \times \frac{15 \times 10^{-6}}{0.0676} = 1.996 \times 10^6 \, \text{N/C} \).

\( 1.9 \times 10^6 \, \text{N/C} \)
\( 1.95 \times 10^6 \, \text{N/C} \)
\( 1.98 \times 10^6 \, \text{N/C} \)
\( 2.0 \times 10^6 \, \text{N/C} \)
4

Three equal charges \( q = 3 \, \mu\text{C} \) are at the vertices of an equilateral triangle of side 1 m. What is the electric field at the centroid?

Distance from vertex to centroid: \( r = \frac{l}{\sqrt{3}} = \frac{1}{\sqrt{3}} \, \text{m} \).

\( E = \frac{k q}{r^2} = 9 \times 10^9 \times \frac{3 \times 10^{-6}}{(1/\sqrt{3})^2} = 81 \times 10^4 \, \text{N/C} \) per charge.

By symmetry (all \( +q \)), vectors cancel, so \( E_{\text{net}} = 0 \).

\( 0 \, \text{N/C} \)
\( 8.1 \times 10^5 \, \text{N/C} \)
\( 1.62 \times 10^6 \, \text{N/C} \)
\( 2.43 \times 10^6 \, \text{N/C} \)
1

Two point charges \( -3 \times 10^{-7} \, \text{C} \) and \( 5 \times 10^{-7} \, \text{C} \) are 80 cm apart in vacuum. What is the magnitude of the force between them?

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

\( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( q_1 = -3 \times 10^{-7} \, \text{C} \), \( q_2 = 5 \times 10^{-7} \, \text{C} \), \( r = 0.8 \, \text{m} \).

\( |q_1 q_2| = 3 \times 5 \times 10^{-14} = 15 \times 10^{-14} \, \text{C}^2 \).

\( r^2 = (0.8)^2 = 0.64 \, \text{m}^2 \).

\( F = 9 \times 10^9 \times \frac{15 \times 10^{-14}}{0.64} = 9 \times 10^9 \times 2.34375 \times 10^{-13} = 0.00211 \, \text{N} \).

0.00211 N
0.0025 N
0.0030 N
0.0035 N
1

A dipole with charges \( +4 \, \mu\text{C} \) and \( -4 \, \mu\text{C} \) separated by 2 mm is in a field \( 7 \times 10^4 \, \text{N/C} \) at 30°. What is the torque?

Dipole moment: \( p = q \times 2a = 4 \times 10^{-6} \times 2 \times 10^{-3} = 8 \times 10^{-9} \, \text{C m} \).

Torque: \( \tau = p E \sin \theta = 8 \times 10^{-9} \times 7 \times 10^4 \times \sin 30^\circ = 5.6 \times 10^{-4} \times 0.5 = 2.8 \times 10^{-4} \, \text{N m} \).

\( 2.5 \times 10^{-4} \, \text{N m} \)
\( 2.8 \times 10^{-4} \, \text{N m} \)
\( 3.0 \times 10^{-4} \, \text{N m} \)
\( 3.2 \times 10^{-4} \, \text{N m} \)
2

A dipole with moment \( p = 5 \times 10^{-9} \, \text{C m} \) is at 60° to a uniform field \( E = 4 \times 10^4 \, \text{N/C} \). What is the torque magnitude?

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

\( \tau = 5 \times 10^{-9} \times 4 \times 10^4 \times \sin 60^\circ = 20 \times 10^{-5} \times \frac{\sqrt{3}}{2} = 1.732 \times 10^{-4} \, \text{N m} \).

\( 1.0 \times 10^{-4} \, \text{N m} \)
\( 1.73 \times 10^{-4} \, \text{N m} \)
\( 2.0 \times 10^{-4} \, \text{N m} \)
\( 2.5 \times 10^{-4} \, \text{N m} \)
2

A dipole with \( p = 7 \times 10^{-9} \, \text{C m} \) is at 45° to a field \( E = 4 \times 10^4 \, \text{N/C} \). What is the torque?

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

\( \tau = 7 \times 10^{-9} \times 4 \times 10^4 \times \sin 45^\circ = 28 \times 10^{-5} \times \frac{\sqrt{2}}{2} = 1.98 \times 10^{-4} \, \text{N m} \).

\( 1.7 \times 10^{-4} \, \text{N m} \)
\( 1.8 \times 10^{-4} \, \text{N m} \)
\( 1.9 \times 10^{-4} \, \text{N m} \)
\( 1.98 \times 10^{-4} \, \text{N m} \)
4

What explains why the electric field due to a dipole can be zero at certain points along its equatorial plane?

In the equatorial plane, the fields from the dipole’s positive and negative charges are equal in magnitude and opposite in direction at the midpoint. Their vector sum cancels out, resulting in a zero field at that specific location.

Charge quantization
Field symmetry
Field cancellation
Charge density
3

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0