Electric Charges and Fields Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Two charges \( +11 \, \mu\text{C} \) and \( -5 \, \mu\text{C} \) are 80 cm apart. What is the electric field magnitude at the midpoint?

Midpoint distance = 40 cm = 0.4 m.

\( E_1 = 9 \times 10^9 \times \frac{11 \times 10^{-6}}{(0.4)^2} = 6.1875 \times 10^5 \, \text{N/C} \) (towards \( -5 \, \mu\text{C} \)).

\( E_2 = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{(0.4)^2} = 2.8125 \times 10^5 \, \text{N/C} \) (towards \( -5 \, \mu\text{C} \)).

Net \( E = 6.1875 \times 10^5 + 2.8125 \times 10^5 = 9 \times 10^5 \, \text{N/C} \).

\( 8.8 \times 10^5 \, \text{N/C} \)
\( 9.0 \times 10^5 \, \text{N/C} \)
\( 9.2 \times 10^5 \, \text{N/C} \)
\( 9.5 \times 10^5 \, \text{N/C} \)
2

A thin spherical shell of radius 24 cm has a charge of \( 16 \, \mu\text{C} \). What is the electric field at a point 30 cm from the center?

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

\( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( q = 16 \times 10^{-6} \, \text{C} \), \( r = 0.3 \, \text{m} \).

\( E = 9 \times 10^9 \times \frac{16 \times 10^{-6}}{(0.3)^2} = 9 \times 10^9 \times \frac{16 \times 10^{-6}}{0.09} = 1.6 \times 10^6 \, \text{N/C} \).

\( 1.6 \times 10^6 \, \text{N/C} \)
\( 1.7 \times 10^6 \, \text{N/C} \)
\( 1.8 \times 10^6 \, \text{N/C} \)
\( 1.9 \times 10^6 \, \text{N/C} \)
1

A conducting sphere of radius 29 cm has a surface charge density of \( 50 \, \mu\text{C/m}^2 \). What is the total charge?

Surface area: \( A = 4 \pi r^2 = 4 \pi (0.29)^2 = 0.335 \pi \, \text{m}^2 \).

Charge: \( q = \sigma A = 50 \times 10^{-6} \times 0.335 \pi = 5.27 \times 10^{-5} \, \text{C} \).

\( 5.0 \times 10^{-5} \, \text{C} \)
\( 5.2 \times 10^{-5} \, \text{C} \)
\( 5.27 \times 10^{-5} \, \text{C} \)
\( 5.3 \times 10^{-5} \, \text{C} \)
3

A dipole with charges \( +3 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) separated by 5 mm is in a field \( 8 \times 10^4 \, \text{N/C} \) at 45°. What is the torque?

Dipole moment: \( p = q \times 2a = 3 \times 10^{-6} \times 5 \times 10^{-3} = 1.5 \times 10^{-8} \, \text{C m} \).

Torque: \( \tau = p E \sin \theta = 1.5 \times 10^{-8} \times 8 \times 10^4 \times \sin 45^\circ = 1.2 \times 10^{-3} \times \frac{\sqrt{2}}{2} = 8.48 \times 10^{-4} \, \text{N m} \).

\( 8.0 \times 10^{-4} \, \text{N m} \)
\( 8.48 \times 10^{-4} \, \text{N m} \)
\( 9.0 \times 10^{-4} \, \text{N m} \)
\( 9.5 \times 10^{-4} \, \text{N m} \)
2

A dipole with charges \( +5 \, \mu\text{C} \) and \( -5 \, \mu\text{C} \) separated by 3 mm is in a field of \( 2 \times 10^4 \, \text{N/C} \) at 45°. What is the torque?

Dipole moment: \( p = q \times 2a = 5 \times 10^{-6} \times 3 \times 10^{-3} = 1.5 \times 10^{-8} \, \text{C m} \).

Torque: \( \tau = p E \sin \theta = 1.5 \times 10^{-8} \times 2 \times 10^4 \times \sin 45^\circ = 3 \times 10^{-4} \times \frac{\sqrt{2}}{2} = 2.12 \times 10^{-4} \, \text{N m} \).

\( 1.8 \times 10^{-4} \, \text{N m} \)
\( 2.12 \times 10^{-4} \, \text{N m} \)
\( 2.5 \times 10^{-4} \, \text{N m} \)
\( 3.0 \times 10^{-4} \, \text{N m} \)
2

What property of insulators causes charges to remain localized when applied to their surface?

Insulators lack free charge carriers (e.g., electrons), preventing charge movement. Applied charges stay where they are placed due to the absence of conductivity, unlike conductors where charges redistribute freely.

High charge density
Lack of free charges
Charge conservation
Field symmetry
2

Three charges \( +5 \, \mu\text{C}, -5 \, \mu\text{C}, +2 \, \mu\text{C} \) are at the vertices of an equilateral triangle of side 1.5 m. What is the force magnitude on \( +5 \, \mu\text{C} \)?

\( F_1 = 9 \times 10^9 \times \frac{5 \times 5 \times 10^{-12}}{(1.5)^2} = 0.1 \, \text{N} \) (attractive).

\( F_2 = 9 \times 10^9 \times \frac{5 \times 2 \times 10^{-12}}{(1.5)^2} = 0.04 \, \text{N} \) (repulsive).

Angle 60°. Net \( F = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos 60^\circ} = \sqrt{0.1^2 + 0.04^2 + 0.004} = 0.129 \, \text{N} \).

0.10 N
0.12 N
0.129 N
0.14 N
3

Three charges \( +5 \, \mu\text{C}, -3 \, \mu\text{C}, +4 \, \mu\text{C} \) are at the vertices of an equilateral triangle of side 1.5 m. What is the force magnitude on \( +5 \, \mu\text{C} \)?

\( F_1 = 9 \times 10^9 \times \frac{5 \times 3 \times 10^{-12}}{(1.5)^2} = 0.06 \, \text{N} \) (attractive).

\( F_2 = 9 \times 10^9 \times \frac{5 \times 4 \times 10^{-12}}{(1.5)^2} = 0.08 \, \text{N} \) (repulsive).

Angle 60°. Net \( F = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos 60^\circ} = \sqrt{0.06^2 + 0.08^2 + 0.0048} = 0.108 \, \text{N} \).

0.09 N
0.10 N
0.108 N
0.12 N
3

Two charges \( +8 \, \mu\text{C} \) and \( -4 \, \mu\text{C} \) are 40 cm apart. What is the electric field magnitude at the midpoint?

Midpoint distance = 20 cm = 0.2 m.

\( E_1 = 9 \times 10^9 \times \frac{8 \times 10^{-6}}{(0.2)^2} = 1.8 \times 10^6 \, \text{N/C} \) (towards \( -4 \, \mu\text{C} \)).

\( E_2 = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{(0.2)^2} = 9 \times 10^5 \, \text{N/C} \) (towards \( -4 \, \mu\text{C} \)).

Net \( E = 1.8 \times 10^6 + 9 \times 10^5 = 2.7 \times 10^6 \, \text{N/C} \).

\( 2.5 \times 10^6 \, \text{N/C} \)
\( 2.7 \times 10^6 \, \text{N/C} \)
\( 2.9 \times 10^6 \, \text{N/C} \)
\( 3.1 \times 10^6 \, \text{N/C} \)
2

Why does the electric field inside a charged conducting shell remain unaffected by charges placed outside it?

Gauss’s law shows that the field inside depends only on enclosed charge. External charges induce surface charges on the conductor, but these adjust to cancel the external field inside, leaving it zero regardless of outside charges.

Charge quantization
Field symmetry
No enclosed charge
Charge mobility
3

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