Electric Charges and Fields Chapter-Wise Test 16

Correct answer Carries: 4.

Wrong Answer Carries: -1.

An infinite line charge has \( \lambda = 5 \times 10^{-7} \, \text{C/m} \). What is the electric field at 10 cm?

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

\( E = \frac{2 \times 9 \times 10^9 \times 5 \times 10^{-7}}{0.1} = 9 \times 10^4 \, \text{N/C} \).

\( 7.5 \times 10^4 \, \text{N/C} \)
\( 8.0 \times 10^4 \, \text{N/C} \)
\( 8.5 \times 10^4 \, \text{N/C} \)
\( 9.0 \times 10^4 \, \text{N/C} \)
4

A dipole with charges \( +6 \, \mu\text{C} \) and \( -6 \, \mu\text{C} \) separated by 2 mm is in a field \( 8 \times 10^4 \, \text{N/C} \) at 60°. What is the torque?

Dipole moment: \( p = q \times 2a = 6 \times 10^{-6} \times 2 \times 10^{-3} = 1.2 \times 10^{-8} \, \text{C m} \).

Torque: \( \tau = p E \sin \theta = 1.2 \times 10^{-8} \times 8 \times 10^4 \times \sin 60^\circ = 9.6 \times 10^{-4} \times \frac{\sqrt{3}}{2} = 8.31 \times 10^{-4} \, \text{N m} \).

\( 8.0 \times 10^{-4} \, \text{N m} \)
\( 8.31 \times 10^{-4} \, \text{N m} \)
\( 8.5 \times 10^{-4} \, \text{N m} \)
\( 9.0 \times 10^{-4} \, \text{N m} \)
2

What causes the electric field to be discontinuous across a charged surface, such as a thin sheet?

The surface charge density on the sheet generates a field that changes abruptly. Gauss’s law shows the field on either side is \( \sigma/2\varepsilon_0 \), with opposite directions, resulting in a discontinuity equal to \( \sigma/\varepsilon_0 \) across the surface.

Charge quantization
Field symmetry
Charge mobility
Surface charge density
4

A thin spherical shell of radius 19 cm has \( q = 7 \, \mu\text{C} \). What is the electric field at 15 cm from the center?

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

\( 1.8 \times 10^6 \, \text{N/C} \)
\( 1.5 \times 10^6 \, \text{N/C} \)
\( 0 \, \text{N/C} \)
\( 1.0 \times 10^6 \, \text{N/C} \)
3

A uniform electric field \( E = 4 \times 10^3 \, \text{N/C} \) is along the x-axis. What is the flux through a square of side 70 cm in the yz-plane?

Area vector \( \Delta \mathbf{S} = (0.7)^2 = 0.49 \, \text{m}^2 \) along x-axis.

Flux: \( \phi = \mathbf{E} \cdot \Delta \mathbf{S} = 4 \times 10^3 \times 0.49 = 1960 \, \text{Nm}^2/\text{C} \).

\( 1960 \, \text{Nm}^2/\text{C} \)
\( 2000 \, \text{Nm}^2/\text{C} \)
\( 2100 \, \text{Nm}^2/\text{C} \)
\( 2200 \, \text{Nm}^2/\text{C} \)
1

A uniform electric field \( E = 6 \times 10^3 \, \text{N/C} \) is along the z-axis. What is the flux through a square of side 50 cm in the xy-plane?

Area vector \( \Delta \mathbf{S} = (0.5)^2 = 0.25 \, \text{m}^2 \) along z-axis.

Flux: \( \phi = \mathbf{E} \cdot \Delta \mathbf{S} = 6 \times 10^3 \times 0.25 = 1500 \, \text{Nm}^2/\text{C} \).

\( 1500 \, \text{Nm}^2/\text{C} \)
\( 1600 \, \text{Nm}^2/\text{C} \)
\( 1700 \, \text{Nm}^2/\text{C} \)
\( 1800 \, \text{Nm}^2/\text{C} \)
1

Which property of electric field lines ensures they never intersect, even in complex charge distributions?

Electric field lines represent the direction of the field at each point. If they intersected, it would imply two different field directions at the same point, which is impossible since the electric field is a unique vector quantity at any given position.

Uniqueness of direction
Continuous nature
Density variation
Symmetry
1

Two charges \( +6 \, \mu\text{C} \) and \( -4 \, \mu\text{C} \) are 25 cm apart. What is the distance from \( +6 \, \mu\text{C} \) where the electric field is zero?

Let distance from \( +6 \, \mu\text{C} \) be \( x \), then from \( -4 \, \mu\text{C} \) is \( 0.25 - x \).

\( \frac{k \times 6 \times 10^{-6}}{x^2} = \frac{k \times 4 \times 10^{-6}}{(0.25 - x)^2} \).

\( \frac{6}{x^2} = \frac{4}{(0.25 - x)^2} \), \( 6 (0.25 - x)^2 = 4 x^2 \).

\( 6 (0.0625 - 0.5 x + x^2) = 4 x^2 \), \( 0.375 - 3 x + 6 x^2 = 4 x^2 \).

\( 2 x^2 - 3 x + 0.375 = 0 \), \( x = \frac{3 \pm \sqrt{9 - 3}}{4} = \frac{3 \pm \sqrt{6}}{4} \).

\( x = 0.15 \, \text{m} \) (between charges).

10 cm
12 cm
13 cm
15 cm
4

A dipole with charges \( +6 \, \mu\text{C} \) and \( -6 \, \mu\text{C} \) separated by 4 mm is in a field \( 5 \times 10^4 \, \text{N/C} \) at 60°. What is the torque?

Dipole moment: \( p = q \times 2a = 6 \times 10^{-6} \times 4 \times 10^{-3} = 2.4 \times 10^{-8} \, \text{C m} \).

Torque: \( \tau = p E \sin \theta = 2.4 \times 10^{-8} \times 5 \times 10^4 \times \sin 60^\circ = 1.2 \times 10^{-3} \times \frac{\sqrt{3}}{2} = 1.04 \times 10^{-3} \, \text{N m} \).

\( 0.9 \times 10^{-3} \, \text{N m} \)
\( 1.04 \times 10^{-3} \, \text{N m} \)
\( 1.2 \times 10^{-3} \, \text{N m} \)
\( 1.5 \times 10^{-3} \, \text{N m} \)
2

A dipole with \( p = 8 \times 10^{-9} \, \text{C m} \) is at 90° to a field \( E = 6 \times 10^4 \, \text{N/C} \). What is the torque?

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

\( \tau = 8 \times 10^{-9} \times 6 \times 10^4 \times \sin 90^\circ = 48 \times 10^{-5} \times 1 = 4.8 \times 10^{-4} \, \text{N m} \).

\( 4.5 \times 10^{-4} \, \text{N m} \)
\( 4.6 \times 10^{-4} \, \text{N m} \)
\( 4.7 \times 10^{-4} \, \text{N m} \)
\( 4.8 \times 10^{-4} \, \text{N m} \)
4

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