Electric Charges and Fields Chapter-Wise Test 14

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A uniform electric field \( E = 2 \times 10^4 \, \text{N/C} \) is along the x-axis. What is the flux through a square of side 20 cm in the yz-plane?

Area vector \( \Delta \mathbf{S} = 0.04 \, \text{m}^2 \) along x-axis.

Flux: \( \phi = \mathbf{E} \cdot \Delta \mathbf{S} = (2 \times 10^4) \times 0.04 = 800 \, \text{Nm}^2/\text{C} \).

\( 400 \, \text{Nm}^2/\text{C} \)
\( 800 \, \text{Nm}^2/\text{C} \)
\( 1200 \, \text{Nm}^2/\text{C} \)
\( 0 \, \text{Nm}^2/\text{C} \)
2

A uniform field \( E = 5 \times 10^3 \, \text{N/C} \) is along the x-axis. What is the flux through a circle of radius 20 cm in the yz-plane?

Area: \( A = \pi (0.2)^2 = 0.1256 \, \text{m}^2 \).

Flux: \( \phi = E A \cos 0^\circ = 5 \times 10^3 \times 0.1256 = 628 \, \text{Nm}^2/\text{C} \).

600 \( \text{Nm}^2/\text{C} \)
620 \( \text{Nm}^2/\text{C} \)
628 \( \text{Nm}^2/\text{C} \)
650 \( \text{Nm}^2/\text{C} \)
3

A body acquires a charge of \( -9.6 \times 10^{-8} \, \text{C} \) when rubbed. How many electrons were transferred to it?

Negative charge indicates electrons gained.

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

\( n = \frac{q}{|e|} = \frac{9.6 \times 10^{-8}}{1.6 \times 10^{-19}} = 6 \times 10^{11} \).

\( 6 \times 10^{11} \)
\( 6.5 \times 10^{11} \)
\( 7 \times 10^{11} \)
\( 7.5 \times 10^{11} \)
1

A conducting sphere of radius 18 cm has a surface charge density of \( 25 \, \mu\text{C/m}^2 \). What is the electric field just outside its surface?

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

\( E = \frac{25 \times 10^{-6}}{8.854 \times 10^{-12}} = 2.82 \times 10^6 \, \text{N/C} \).

\( 2.5 \times 10^6 \, \text{N/C} \)
\( 2.7 \times 10^6 \, \text{N/C} \)
\( 2.82 \times 10^6 \, \text{N/C} \)
\( 3.0 \times 10^6 \, \text{N/C} \)
3

An infinite line charge produces an electric field of \( 1.8 \times 10^5 \, \text{N/C} \) at 3 cm. What is its linear charge density?

\( E = \frac{\lambda}{2 \pi \varepsilon_0 r} \), or \( E = \frac{2 k \lambda}{r} \), \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \).

\( 1.8 \times 10^5 = \frac{2 \times 9 \times 10^9 \times \lambda}{0.03} \).

\( \lambda = \frac{1.8 \times 10^5 \times 0.03}{18 \times 10^9} = 3 \times 10^{-7} \, \text{C/m} \).

\( 2.5 \times 10^{-7} \, \text{C/m} \)
\( 3.0 \times 10^{-7} \, \text{C/m} \)
\( 3.5 \times 10^{-7} \, \text{C/m} \)
\( 4.0 \times 10^{-7} \, \text{C/m} \)
2

Three charges \( +4 \, \mu\text{C}, -4 \, \mu\text{C}, +2 \, \mu\text{C} \) are at the vertices of an equilateral triangle of side 1 m. What is the force magnitude on \( +4 \, \mu\text{C} \)?

\( F_1 = 9 \times 10^9 \times \frac{4 \times 4 \times 10^{-12}}{1} = 0.144 \, \text{N} \) (attractive).

\( F_2 = 9 \times 10^9 \times \frac{4 \times 2 \times 10^{-12}}{1} = 0.072 \, \text{N} \) (repulsive).

Angle 60°. Net \( F = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos 60^\circ} = \sqrt{0.144^2 + 0.072^2 + 0.010368} = 0.187 \, \text{N} \).

0.15 N
0.17 N
0.187 N
0.20 N
3

What ensures that the electric field lines from a positive point charge always point radially outward?

The positive charge generates a repulsive force on a positive test charge, as per Coulomb’s law. Due to the spherical symmetry of a point charge, the field is directed radially outward, reflecting the repulsion and symmetry of the charge distribution.

Charge quantization
Repulsion
Field superposition
Charge density
2

Why does the electric field due to a dipole exhibit a stronger angular dependence compared to a single point charge?

The dipole’s field arises from two opposite charges, creating a directional dependence. The angle between the dipole axis and the point affects the field’s magnitude and direction more significantly than a point charge’s isotropic \( 1/r^2 \) field.

Charge symmetry
Directional dependence
Field cancellation
Charge density
2

Why does the electric field due to a charged conducting sphere remain constant just outside its surface regardless of its radius?

The field just outside a conductor is \( \sigma/\varepsilon_0 \), where \( \sigma \) is the surface charge density. For a sphere, \( \sigma = Q/(4\pi r^2) \), but the field depends only on \( \sigma \) at the surface, not \( r \), due to equilibrium conditions.

Charge quantization
Field symmetry
Surface charge density
Charge mobility
3

A conducting sphere of radius 15 cm has an electric field of \( 4 \times 10^3 \, \text{N/C} \) at 30 cm from its center. What is the charge?

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

\( 4 \times 10^3 = 9 \times 10^9 \times \frac{q}{(0.3)^2} \).

\( q = \frac{4 \times 10^3 \times 0.09}{9 \times 10^9} = 4 \times 10^{-8} \, \text{C} \).

\( 3.5 \times 10^{-8} \, \text{C} \)
\( 3.8 \times 10^{-8} \, \text{C} \)
\( 3.9 \times 10^{-8} \, \text{C} \)
\( 4.0 \times 10^{-8} \, \text{C} \)
4

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