Electric Charges and Fields Chapter-Wise Test 8

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A uniform electric field \( E = 3 \times 10^3 \, \text{N/C} \) is along the y-axis. What is the flux through a square of side 60 cm in the xz-plane?

Area vector \( \Delta \mathbf{S} = (0.6)^2 = 0.36 \, \text{m}^2 \) along y-axis.

Flux: \( \phi = \mathbf{E} \cdot \Delta \mathbf{S} = 3 \times 10^3 \times 0.36 = 1080 \, \text{Nm}^2/\text{C} \).

\( 1080 \, \text{Nm}^2/\text{C} \)
\( 1100 \, \text{Nm}^2/\text{C} \)
\( 1200 \, \text{Nm}^2/\text{C} \)
\( 1300 \, \text{Nm}^2/\text{C} \)
1

A charge of \( 13 \, \mu\text{C} \) is at the center of a cube of edge 50 cm. What is the flux through one face?

Total flux: \( \phi = \frac{q}{\varepsilon_0} = \frac{13 \times 10^{-6}}{8.854 \times 10^{-12}} = 1.468 \times 10^6 \, \text{Nm}^2/\text{C} \).

Flux per face (6 faces): \( \phi_{\text{face}} = \frac{1.468 \times 10^6}{6} = 2.447 \times 10^5 \, \text{Nm}^2/\text{C} \).

\( 2.3 \times 10^5 \, \text{Nm}^2/\text{C} \)
\( 2.45 \times 10^5 \, \text{Nm}^2/\text{C} \)
\( 2.5 \times 10^5 \, \text{Nm}^2/\text{C} \)
\( 2.6 \times 10^5 \, \text{Nm}^2/\text{C} \)
2

A thin spherical shell of radius 6 cm has \( q = 5 \, \mu\text{C} \). What is the electric field at 8 cm from the center?

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

\( E = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{(0.08)^2} = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{0.0064} = 7.03 \times 10^6 \, \text{N/C} \).

\( 6.5 \times 10^6 \, \text{N/C} \)
\( 6.8 \times 10^6 \, \text{N/C} \)
\( 7.0 \times 10^6 \, \text{N/C} \)
\( 7.03 \times 10^6 \, \text{N/C} \)
4

Why does the electric field inside a charged spherical shell remain zero even if the shell is irregular but closed?

Gauss’s law applies to any closed surface: if no charge is enclosed within the shell, the net flux through a Gaussian surface inside is zero. For a conductor, charges reside on the outer surface, ensuring no field inside, regardless of shape.

Charge quantization
Field symmetry
Charge mobility
No enclosed charge
4

A thin spherical shell of radius 5 cm has \( q = 2 \, \mu\text{C} \). What is the electric field at 3 cm from the center?

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

\( 1.0 \times 10^6 \, \text{N/C} \)
\( 0.5 \times 10^6 \, \text{N/C} \)
\( 0 \, \text{N/C} \)
\( 2.0 \times 10^6 \, \text{N/C} \)
3

A point charge \( -8 \, \mu\text{C} \) is at the origin. What is the electric field magnitude at a point 4 m along the z-axis?

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

\( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( q = 8 \times 10^{-6} \, \text{C} \), \( r = 4 \, \text{m} \).

\( E = 9 \times 10^9 \times \frac{8 \times 10^{-6}}{(4)^2} = 9 \times 10^9 \times \frac{8 \times 10^{-6}}{16} = 4.5 \times 10^3 \, \text{N/C} \).

\( 4.5 \times 10^3 \, \text{N/C} \)
\( 4.8 \times 10^3 \, \text{N/C} \)
\( 5.0 \times 10^3 \, \text{N/C} \)
\( 5.2 \times 10^3 \, \text{N/C} \)
1

Why is the electric field inside a conductor zero when it is placed in an external electric field?

In a conductor, free charges redistribute in response to an external field until the internal field cancels the external field completely. This equilibrium condition ensures no net field exists inside, as any field would cause charge motion, violating static conditions.

Charge quantization
High charge density
Charge redistribution
Field symmetry
3

Why does the electric field strength near a charged object increase as the density of field lines increases?

The density of electric field lines represents the magnitude of the field. Closer lines indicate a stronger field because the field strength is proportional to the number of lines per unit area, a convention derived from the inverse-square law and field intensity.

Charge quantization
Field symmetry
Line density
Charge mobility
3

An infinite line charge has \( \lambda = 6 \times 10^{-6} \, \text{C/m} \). What is the electric field at 12 cm?

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

\( E = \frac{2 \times 9 \times 10^9 \times 6 \times 10^{-6}}{0.12} = 9 \times 10^5 \, \text{N/C} \).

\( 8.5 \times 10^5 \, \text{N/C} \)
\( 9.0 \times 10^5 \, \text{N/C} \)
\( 9.5 \times 10^5 \, \text{N/C} \)
\( 1.0 \times 10^6 \, \text{N/C} \)
2

Two charges \( +6 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) are 50 cm apart. What is the electric field magnitude at the midpoint?

Midpoint distance = 25 cm = 0.25 m.

\( E_1 = 9 \times 10^9 \times \frac{6 \times 10^{-6}}{(0.25)^2} = 8.64 \times 10^5 \, \text{N/C} \) (towards \( -3 \, \mu\text{C} \)).

\( E_2 = 9 \times 10^9 \times \frac{3 \times 10^{-6}}{(0.25)^2} = 4.32 \times 10^5 \, \text{N/C} \) (towards \( -3 \, \mu\text{C} \)).

Net \( E = 8.64 \times 10^5 + 4.32 \times 10^5 = 1.296 \times 10^6 \, \text{N/C} \).

\( 1.2 \times 10^6 \, \text{N/C} \)
\( 1.296 \times 10^6 \, \text{N/C} \)
\( 1.4 \times 10^6 \, \text{N/C} \)
\( 1.5 \times 10^6 \, \text{N/C} \)
2

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