Electric Charges and Fields Chapter-Wise Test 11

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A conducting sphere of radius 26 cm has a surface charge density of \( 45 \, \mu\text{C/m}^2 \). What is the electric field just outside its surface?

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

\( E = \frac{45 \times 10^{-6}}{8.854 \times 10^{-12}} = 5.08 \times 10^6 \, \text{N/C} \).

\( 4.8 \times 10^6 \, \text{N/C} \)
\( 5.0 \times 10^6 \, \text{N/C} \)
\( 5.08 \times 10^6 \, \text{N/C} \)
\( 5.2 \times 10^6 \, \text{N/C} \)
3

A point charge \( -10 \, \mu\text{C} \) is at the origin. What is the electric field magnitude at a point 10 m along the x-axis?

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

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

\( E = 9 \times 10^9 \times \frac{10 \times 10^{-6}}{(10)^2} = 9 \times 10^9 \times \frac{10 \times 10^{-6}}{100} = 9 \times 10^2 \, \text{N/C} \).

\( 900 \, \text{N/C} \)
\( 950 \, \text{N/C} \)
\( 1000 \, \text{N/C} \)
\( 1050 \, \text{N/C} \)
1

Why does the electric field outside a charged conducting sphere resemble that of a point charge located at its center?

The spherical symmetry of the charge distribution on the conductor’s surface ensures that the field outside behaves as if all charge were at the center, as per Gauss’s law. This symmetry simplifies the field to a radial, point-charge-like pattern.

Charge quantization
Field cancellation
Charge density
Spherical symmetry
4

A conducting sphere of radius 22 cm has a surface charge density of \( 40 \, \mu\text{C/m}^2 \). What is the electric field just outside its surface?

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

\( E = \frac{40 \times 10^{-6}}{8.854 \times 10^{-12}} = 4.52 \times 10^6 \, \text{N/C} \).

\( 4.2 \times 10^6 \, \text{N/C} \)
\( 4.4 \times 10^6 \, \text{N/C} \)
\( 4.52 \times 10^6 \, \text{N/C} \)
\( 4.7 \times 10^6 \, \text{N/C} \)
3

A plane sheet has \( \sigma = 1.77 \times 10^{-11} \, \text{C/m}^2 \). What is the electric field near it?

\( E = \frac{\sigma}{2 \varepsilon_0} \).

\( E = \frac{1.77 \times 10^{-11}}{2 \times 8.854 \times 10^{-12}} = 1 \, \text{N/C} \).

\( 0.8 \, \text{N/C} \)
\( 1.0 \, \text{N/C} \)
\( 1.2 \, \text{N/C} \)
\( 1.5 \, \text{N/C} \)
2

A dipole with \( p = 8 \times 10^{-9} \, \text{C m} \) is at 30° to a field \( E = 5 \times 10^4 \, \text{N/C} \). What is the torque magnitude?

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

\( \tau = 8 \times 10^{-9} \times 5 \times 10^4 \times \sin 30^\circ = 8 \times 10^{-9} \times 5 \times 10^4 \times 0.5 = 2 \times 10^{-4} \, \text{N m} \).

\( 1.5 \times 10^{-4} \, \text{N m} \)
\( 1.8 \times 10^{-4} \, \text{N m} \)
\( 1.9 \times 10^{-4} \, \text{N m} \)
\( 2.0 \times 10^{-4} \, \text{N m} \)
4

A thin spherical shell of radius 13 cm has a charge of \( 9 \, \mu\text{C} \). What is the electric field at a point 18 cm from the center?

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

\( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( q = 9 \times 10^{-6} \, \text{C} \), \( r = 0.18 \, \text{m} \).

\( E = 9 \times 10^9 \times \frac{9 \times 10^{-6}}{(0.18)^2} = 9 \times 10^9 \times \frac{9 \times 10^{-6}}{0.0324} = 2.5 \times 10^6 \, \text{N/C} \).

\( 2.5 \times 10^6 \, \text{N/C} \)
\( 2.7 \times 10^6 \, \text{N/C} \)
\( 2.9 \times 10^6 \, \text{N/C} \)
\( 3.1 \times 10^6 \, \text{N/C} \)
1

An infinite line charge has \( E = 4.5 \times 10^5 \, \text{N/C} \) at 8 cm. What is \( \lambda \)?

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

\( 4.5 \times 10^5 = \frac{2 \times 9 \times 10^9 \times \lambda}{0.08} \).

\( \lambda = \frac{4.5 \times 10^5 \times 0.08}{18 \times 10^9} = 2 \times 10^{-6} \, \text{C/m} \).

\( 1.8 \times 10^{-6} \, \text{C/m} \)
\( 1.9 \times 10^{-6} \, \text{C/m} \)
\( 1.95 \times 10^{-6} \, \text{C/m} \)
\( 2.0 \times 10^{-6} \, \text{C/m} \)
4

An infinite line charge has a linear charge density of \( 4 \times 10^{-6} \, \text{C/m} \). What is the electric field at a distance of 8 cm?

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

\( E = \frac{2 \times 9 \times 10^9 \times 4 \times 10^{-6}}{0.08} = 9 \times 10^5 \, \text{N/C} \).

\( 8.0 \times 10^5 \, \text{N/C} \)
\( 9.0 \times 10^5 \, \text{N/C} \)
\( 1.0 \times 10^6 \, \text{N/C} \)
\( 1.2 \times 10^6 \, \text{N/C} \)
2

Why does the electric field inside a charged conducting shell remain zero even when an external field is applied?

Charges on the conductor’s surface redistribute to cancel any external field inside, creating an electrostatic shield. This shielding effect ensures the internal field is zero, as charges adjust to maintain equilibrium within the conductor.

Charge quantization
Field symmetry
Electrostatic shielding
Charge density
3

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