Electric Charges and Fields Chapter-Wise Test 7

Correct answer Carries: 4.

Wrong Answer Carries: -1.

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

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

\( \tau = 7 \times 10^{-9} \times 9 \times 10^4 \times \sin 45^\circ = 63 \times 10^{-5} \times \frac{\sqrt{2}}{2} = 4.45 \times 10^{-4} \, \text{N m} \).

\( 4.2 \times 10^{-4} \, \text{N m} \)
\( 4.3 \times 10^{-4} \, \text{N m} \)
\( 4.4 \times 10^{-4} \, \text{N m} \)
\( 4.45 \times 10^{-4} \, \text{N m} \)
4

A thin spherical shell of radius 20 cm has a charge of \( 14 \, \mu\text{C} \). What is the electric field at a point 25 cm from the center?

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

\( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( q = 14 \times 10^{-6} \, \text{C} \), \( r = 0.25 \, \text{m} \).

\( E = 9 \times 10^9 \times \frac{14 \times 10^{-6}}{(0.25)^2} = 9 \times 10^9 \times \frac{14 \times 10^{-6}}{0.0625} = 2.016 \times 10^6 \, \text{N/C} \).

\( 2.016 \times 10^6 \, \text{N/C} \)
\( 2.1 \times 10^6 \, \text{N/C} \)
\( 2.2 \times 10^6 \, \text{N/C} \)
\( 2.3 \times 10^6 \, \text{N/C} \)
1

Why does the electric field due to a uniformly charged wire drop off more slowly with distance compared to a point charge?

The wire’s linear charge distribution leads to a cylindrical field pattern (\( 1/r \)), as per Gauss’s law, spreading over a surface area proportional to \( r \). A point charge’s spherical spread (\( 1/r^2 \)) over \( r^2 \) results in a faster drop-off.

Charge quantization
Field symmetry
Linear distribution
Charge density
3

A charge of \( 4 \, \mu\text{C} \) is placed at the center of a cube of edge 10 cm. What is the electric flux through one face?

Total flux: \( \phi = \frac{q}{\varepsilon_0} = \frac{4 \times 10^{-6}}{8.854 \times 10^{-12}} = 4.52 \times 10^5 \, \text{Nm}^2/\text{C} \).

Flux through one face (cube has 6 faces): \( \phi_{\text{face}} = \frac{\phi}{6} = \frac{4.52 \times 10^5}{6} = 7.53 \times 10^4 \, \text{Nm}^2/\text{C} \).

\( 5.0 \times 10^4 \, \text{Nm}^2/\text{C} \)
\( 7.5 \times 10^4 \, \text{Nm}^2/\text{C} \)
\( 9.0 \times 10^4 \, \text{Nm}^2/\text{C} \)
\( 1.2 \times 10^5 \, \text{Nm}^2/\text{C} \)
2

Three charges \( +q, +q, -q \) are at the vertices of an equilateral triangle of side 3 m. What is the net force magnitude on \( +q \) at one vertex?

Force between \( +q \) and \( +q \): \( F = \frac{k q^2}{(3)^2} = \frac{k q^2}{9} \) (repulsive).

Force between \( +q \) and \( -q \): \( F = \frac{k q^2}{9} \) (attractive).

Angle between forces is 60°. Resultant: \( F_{\text{net}} = \sqrt{F^2 + F^2 + 2F^2 \cos 60^\circ} = \sqrt{3} F = \frac{\sqrt{3} k q^2}{9} \).

\( \frac{k q^2}{9} \)
\( \frac{\sqrt{3} k q^2}{9} \)
\( \frac{2 k q^2}{9} \)
\( \frac{k q^2}{3} \)
2

Which principle explains why rubbing two neutral objects can result in one becoming positively charged and the other negatively charged?

The conservation of charge ensures that the total charge remains zero. Rubbing transfers electrons from one object to another due to differing electron affinities, leaving one with a net positive charge (electron loss) and the other negative (electron gain).

Conservation of charge
Charge quantization
Superposition principle
Field symmetry
1

Two charges \( +14 \, \mu\text{C} \) and \( -6 \, \mu\text{C} \) are 100 cm apart. What is the distance from \( +14 \, \mu\text{C} \) where the electric field is zero?

Let \( x \) be distance from \( +14 \, \mu\text{C} \), then \( 1 - x \) from \( -6 \, \mu\text{C} \).

\( \frac{14 \times 10^{-6}}{x^2} = \frac{6 \times 10^{-6}}{(1 - x)^2} \), \( 14 (1 - x)^2 = 6 x^2 \).

\( 14 (1 - 2 x + x^2) = 6 x^2 \), \( 14 - 28 x + 14 x^2 = 6 x^2 \).

\( 8 x^2 - 28 x + 14 = 0 \), \( 4 x^2 - 14 x + 7 = 0 \), \( x = \frac{14 \pm \sqrt{196 - 112}}{8} = \frac{14 \pm 9.17}{8} \).

\( x = 0.583 \, \text{m} \) (between charges).

56 cm
57 cm
58 cm
58.3 cm
4

What allows the electric field to be discontinuous across a charged surface, such as a conductor’s boundary?

Surface charge density creates a field discontinuity. Inside a conductor, the field is zero, while just outside, it’s proportional to the surface charge density (\( E = \sigma/\varepsilon_0 \)), as per Gauss’s law applied to a pillbox surface straddling the boundary.

Charge quantization
Field symmetry
Charge mobility
Surface charge density
4

A conducting sphere of radius 10 cm has a surface charge density of \( 20 \, \mu\text{C/m}^2 \). What is the electric field just outside its surface?

For a conductor, \( E = \frac{\sigma}{\varepsilon_0} \).

\( E = \frac{20 \times 10^{-6}}{8.854 \times 10^{-12}} = 2.26 \times 10^6 \, \text{N/C} \).

\( 1.8 \times 10^6 \, \text{N/C} \)
\( 2.0 \times 10^6 \, \text{N/C} \)
\( 2.26 \times 10^6 \, \text{N/C} \)
\( 2.5 \times 10^6 \, \text{N/C} \)
3

Two charges \( +12 \, \mu\text{C} \) and \( -4 \, \mu\text{C} \) are 90 cm apart. What is the distance from \( +12 \, \mu\text{C} \) where the electric field is zero?

Let \( x \) be distance from \( +12 \, \mu\text{C} \), then \( 0.9 - x \) from \( -4 \, \mu\text{C} \).

\( \frac{12 \times 10^{-6}}{x^2} = \frac{4 \times 10^{-6}}{(0.9 - x)^2} \), \( 12 (0.9 - x)^2 = 4 x^2 \).

\( 3 (0.81 - 1.8 x + x^2) = x^2 \), \( 2.43 - 5.4 x + 3 x^2 = x^2 \).

\( 2 x^2 - 5.4 x + 2.43 = 0 \), \( x = \frac{5.4 \pm \sqrt{29.16 - 19.44}}{4} = \frac{5.4 \pm 3.12}{4} \).

\( x = 0.57 \, \text{m} \) (between charges).

55 cm
56 cm
56.5 cm
57 cm
4

Two charges \( +3 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) are 15 cm apart. What is the force on a \( 2 \, \mu\text{C} \) charge placed at the midpoint?

Electric field at midpoint: \( E_1 = 9 \times 10^9 \times \frac{3 \times 10^{-6}}{(0.075)^2} = 4.8 \times 10^6 \, \text{N/C} \) (towards \( -3 \, \mu\text{C} \)).

\( E_2 = 4.8 \times 10^6 \, \text{N/C} \) (towards \( -3 \, \mu\text{C} \)).

Net \( E = 4.8 \times 10^6 + 4.8 \times 10^6 = 9.6 \times 10^6 \, \text{N/C} \).

Force: \( F = q E = 2 \times 10^{-6} \times 9.6 \times 10^6 = 19.2 \, \text{N} \).

15 N
17.5 N
19.2 N
21 N
3

An infinite line charge has \( \lambda = 8 \times 10^{-7} \, \text{C/m} \). What is the electric field at 20 cm?

\( E = \frac{2 k \lambda}{r} \), \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \).

\( E = \frac{2 \times 9 \times 10^9 \times 8 \times 10^{-7}}{0.2} = 7.2 \times 10^4 \, \text{N/C} \).

\( 6.5 \times 10^4 \, \text{N/C} \)
\( 7.2 \times 10^4 \, \text{N/C} \)
\( 7.5 \times 10^4 \, \text{N/C} \)
\( 8.0 \times 10^4 \, \text{N/C} \)
2

A charge of \( 8 \, \mu\text{C} \) is at the center of a cube of edge 25 cm. What is the total flux through the cube?

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

\( \phi = \frac{8 \times 10^{-6}}{8.854 \times 10^{-12}} = 9.03 \times 10^5 \, \text{Nm}^2/\text{C} \).

\( 8.5 \times 10^5 \, \text{Nm}^2/\text{C} \)
\( 9.03 \times 10^5 \, \text{Nm}^2/\text{C} \)
\( 9.5 \times 10^5 \, \text{Nm}^2/\text{C} \)
\( 1.0 \times 10^6 \, \text{Nm}^2/\text{C} \)
2

A uniform field \( E = 9 \times 10^3 \, \text{N/C} \) is along the x-axis. What is the flux through a rectangle of 15 cm × 20 cm in the yz-plane?

Area: \( A = 0.15 \times 0.2 = 0.03 \, \text{m}^2 \).

Flux: \( \phi = E A \cos 0^\circ = 9 \times 10^3 \times 0.03 = 270 \, \text{Nm}^2/\text{C} \).

250 \( \text{Nm}^2/\text{C} \)
260 \( \text{Nm}^2/\text{C} \)
270 \( \text{Nm}^2/\text{C} \)
280 \( \text{Nm}^2/\text{C} \)
3

A dipole with \( p = 5 \times 10^{-9} \, \text{C m} \) is at 90° to a field \( E = 5 \times 10^4 \, \text{N/C} \). What is the torque?

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

\( \tau = 5 \times 10^{-9} \times 5 \times 10^4 \times \sin 90^\circ = 25 \times 10^{-5} \times 1 = 2.5 \times 10^{-4} \, \text{N m} \).

\( 2.2 \times 10^{-4} \, \text{N m} \)
\( 2.3 \times 10^{-4} \, \text{N m} \)
\( 2.4 \times 10^{-4} \, \text{N m} \)
\( 2.5 \times 10^{-4} \, \text{N m} \)
4

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