Electric Charges and Fields Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A plane sheet has \( \sigma = 5.31 \times 10^{-11} \, \text{C/m}^2 \). What is the electric field near it?

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

\( E = \frac{5.31 \times 10^{-11}}{2 \times 8.854 \times 10^{-12}} = 3 \, \text{N/C} \).

\( 2.5 \, \text{N/C} \)
\( 3.0 \, \text{N/C} \)
\( 3.5 \, \text{N/C} \)
\( 4.0 \, \text{N/C} \)
2

Two charges \( +5 \, \mu\text{C} \) and \( -5 \, \mu\text{C} \) are 30 cm apart. What is the potential energy of the system?

Potential energy: \( U = k \frac{q_1 q_2}{r} \).

\( U = 9 \times 10^9 \times \frac{(5 \times 10^{-6}) \times (-5 \times 10^{-6})}{0.3} = 9 \times 10^9 \times \frac{-25 \times 10^{-12}}{0.3} = -0.75 \, \text{J} \).

\( -0.9 \, \text{J} \)
\( -0.75 \, \text{J} \)
\( -0.6 \, \text{J} \)
\( 0.75 \, \text{J} \)
2

How many electrons must be added to a neutral body to give it a charge of \( -8 \times 10^{-8} \, \text{C} \)?

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

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

\( 4 \times 10^{11} \)
\( 5 \times 10^{11} \)
\( 6 \times 10^{11} \)
\( 7 \times 10^{11} \)
2

A charge of \( 5 \, \mu\text{C} \) is enclosed in a cube of edge 30 cm. What is the flux through one face?

Total flux: \( \phi = \frac{q}{\varepsilon_0} = \frac{5 \times 10^{-6}}{8.854 \times 10^{-12}} = 5.65 \times 10^5 \, \text{Nm}^2/\text{C} \).

Flux per face (6 faces): \( \phi_{\text{face}} = \frac{5.65 \times 10^5}{6} = 9.42 \times 10^4 \, \text{Nm}^2/\text{C} \).

\( 9.0 \times 10^4 \, \text{Nm}^2/\text{C} \)
\( 9.42 \times 10^4 \, \text{Nm}^2/\text{C} \)
\( 9.8 \times 10^4 \, \text{Nm}^2/\text{C} \)
\( 1.0 \times 10^5 \, \text{Nm}^2/\text{C} \)
2

A dipole with charges \( +2 \, \mu\text{C} \) and \( -2 \, \mu\text{C} \) separated by 1 mm is in a field of \( 6 \times 10^4 \, \text{N/C} \) at 90°. What is the torque?

Dipole moment: \( p = q \times 2a = 2 \times 10^{-6} \times 1 \times 10^{-3} = 2 \times 10^{-9} \, \text{C m} \).

Torque: \( \tau = p E \sin \theta = 2 \times 10^{-9} \times 6 \times 10^4 \times \sin 90^\circ = 1.2 \times 10^{-4} \, \text{N m} \).

\( 1.0 \times 10^{-4} \, \text{N m} \)
\( 1.2 \times 10^{-4} \, \text{N m} \)
\( 1.5 \times 10^{-4} \, \text{N m} \)
\( 1.8 \times 10^{-4} \, \text{N m} \)
2

A charge of \( -2 \, \mu\text{C} \) produces a flux of \( -2.26 \times 10^5 \, \text{Nm}^2/\text{C} \) through a spherical surface. What is the radius of the surface?

Flux: \( \phi = \frac{q}{\varepsilon_0} \), \( q = -2 \times 10^{-6} \, \text{C} \), \( \varepsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \).

\( \phi = \frac{-2 \times 10^{-6}}{8.854 \times 10^{-12}} = -2.26 \times 10^5 \, \text{Nm}^2/\text{C} \), matches given flux, so radius is irrelevant (total flux is constant).

Question likely tests understanding; flux is independent of radius.

10 cm
20 cm
Flux is independent of radius
50 cm
3

What fundamental concept explains why the electric field just outside a charged conductor is perpendicular to its surface?

In electrostatic equilibrium, the electric field inside a conductor is zero. If the field just outside had a tangential component, charges would move along the surface, contradicting equilibrium. Thus, the field must be normal to the surface.

Charge quantization
Field symmetry
Electrostatic equilibrium
Superposition principle
3

Why does the electric field inside a charged insulator depend on the distribution of charges within it?

Fixed charges in an insulator create a field based on their spatial arrangement. Gauss’s law shows the field at a point depends on the enclosed charge, which varies with position due to the immovable nature of the charges.

Charge quantization
Field symmetry
Charge mobility
Charge distribution
4

A uniform field \( E = 8 \times 10^3 \, \text{N/C} \) is along the y-axis. What is the flux through a circle of radius 15 cm in the xz-plane at 30° to the y-axis?

Area: \( A = \pi (0.15)^2 = 0.0707 \, \text{m}^2 \).

Flux: \( \phi = E A \cos 30^\circ = 8 \times 10^3 \times 0.0707 \times \frac{\sqrt{3}}{2} = 489.7 \, \text{Nm}^2/\text{C} \).

450 \( \text{Nm}^2/\text{C} \)
470 \( \text{Nm}^2/\text{C} \)
490 \( \text{Nm}^2/\text{C} \)
510 \( \text{Nm}^2/\text{C} \)
3

A thin spherical shell of radius 30 cm has a charge of \( 18 \, \mu\text{C} \). What is the electric field at a point 36 cm from the center?

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

\( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( q = 18 \times 10^{-6} \, \text{C} \), \( r = 0.36 \, \text{m} \).

\( E = 9 \times 10^9 \times \frac{18 \times 10^{-6}}{(0.36)^2} = 9 \times 10^9 \times \frac{18 \times 10^{-6}}{0.1296} = 1.25 \times 10^6 \, \text{N/C} \).

\( 1.25 \times 10^6 \, \text{N/C} \)
\( 1.3 \times 10^6 \, \text{N/C} \)
\( 1.4 \times 10^6 \, \text{N/C} \)
\( 1.5 \times 10^6 \, \text{N/C} \)
1

An infinite line charge has a linear charge density \( \lambda = 2 \times 10^{-6} \, \text{C/m} \). What is the electric field at 5 cm from the wire?

\( E = \frac{\lambda}{2 \pi \varepsilon_0 r} \), \( k = \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \), so \( \frac{1}{\varepsilon_0} = 4 \pi \times 9 \times 10^9 \).

\( E = \frac{2 \times 10^{-6}}{2 \pi \times 8.854 \times 10^{-12} \times 0.05} = 7.19 \times 10^5 \, \text{N/C} \).

Or, \( E = \frac{2k\lambda}{r} = \frac{2 \times 9 \times 10^9 \times 2 \times 10^{-6}}{0.05} = 7.2 \times 10^5 \, \text{N/C} \).

\( 5.4 \times 10^5 \, \text{N/C} \)
\( 7.2 \times 10^5 \, \text{N/C} \)
\( 9.0 \times 10^5 \, \text{N/C} \)
\( 1.08 \times 10^6 \, \text{N/C} \)
2

A conducting sphere of radius 14 cm has an electric field of \( 7 \times 10^3 \, \text{N/C} \) at 28 cm from its center. What is the charge?

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

\( 7 \times 10^3 = 9 \times 10^9 \times \frac{q}{(0.28)^2} \).

\( q = \frac{7 \times 10^3 \times 0.0784}{9 \times 10^9} = 6.1 \times 10^{-8} \, \text{C} \).

\( 5.8 \times 10^{-8} \, \text{C} \)
\( 5.9 \times 10^{-8} \, \text{C} \)
\( 6.0 \times 10^{-8} \, \text{C} \)
\( 6.1 \times 10^{-8} \, \text{C} \)
4

Three charges \( +7 \, \mu\text{C}, -5 \, \mu\text{C}, +3 \, \mu\text{C} \) are at the vertices of an equilateral triangle of side 2 m. What is the force magnitude on \( +7 \, \mu\text{C} \)?

\( F_1 = 9 \times 10^9 \times \frac{7 \times 5 \times 10^{-12}}{(2)^2} = 0.07875 \, \text{N} \) (attractive).

\( F_2 = 9 \times 10^9 \times \frac{7 \times 3 \times 10^{-12}}{(2)^2} = 0.04725 \, \text{N} \) (repulsive).

Angle 60°. Net \( F = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos 60^\circ} = \sqrt{0.07875^2 + 0.04725^2 + 0.00372} = 0.098 \, \text{N} \).

0.09 N
0.095 N
0.098 N
0.10 N
3

A uniform field \( E = 5 \times 10^3 \, \text{N/C} \) is along the x-axis. What is the net flux through a cube of side 30 cm with faces parallel to coordinate planes?

Flux through face at \( x = 0 \): \( \phi = E \times A = 5 \times 10^3 \times (0.3)^2 = 450 \, \text{Nm}^2/\text{C} \) (inward).

Flux through face at \( x = 0.3 \): \( 450 \, \text{Nm}^2/\text{C} \) (outward).

Net flux: \( 450 - 450 = 0 \, \text{Nm}^2/\text{C} \) (no charge enclosed).

\( 0 \, \text{Nm}^2/\text{C} \)
\( 450 \, \text{Nm}^2/\text{C} \)
\( 900 \, \text{Nm}^2/\text{C} \)
\( 1350 \, \text{Nm}^2/\text{C} \)
1

A thin spherical shell of radius 11 cm has \( q = 5 \, \mu\text{C} \). What is the electric field at 9 cm from the center?

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

\( 1.2 \times 10^6 \, \text{N/C} \)
\( 1.0 \times 10^6 \, \text{N/C} \)
\( 0 \, \text{N/C} \)
\( 0.8 \times 10^6 \, \text{N/C} \)
3

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