Electric Charges and Fields Chapter-Wise Test 19

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A plane sheet has \( \sigma = 1.416 \times 10^{-10} \, \text{C/m}^2 \). What is the electric field near it?

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

\( E = \frac{1.416 \times 10^{-10}}{2 \times 8.854 \times 10^{-12}} = 8 \, \text{N/C} \).

\( 7.5 \, \text{N/C} \)
\( 8.0 \, \text{N/C} \)
\( 8.5 \, \text{N/C} \)
\( 9.0 \, \text{N/C} \)
2

Why does Gauss’s law fail to determine the electric field for a finite charged object without symmetry?

Gauss’s law requires a Gaussian surface with symmetry matching the charge distribution to simplify field calculation. For finite, asymmetric objects, the field varies in complex ways, making symmetry-based simplification impossible without additional methods.

Charge quantization
Field superposition
Lack of charge
Lack of symmetry
4

A plane sheet has \( \sigma = 7.08 \times 10^{-11} \, \text{C/m}^2 \). What is the electric field near it?

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

\( E = \frac{7.08 \times 10^{-11}}{2 \times 8.854 \times 10^{-12}} = 4 \, \text{N/C} \).

\( 3.5 \, \text{N/C} \)
\( 4.0 \, \text{N/C} \)
\( 4.5 \, \text{N/C} \)
\( 5.0 \, \text{N/C} \)
2

A conducting sphere of radius 12 cm has an electric field of \( 6 \times 10^3 \, \text{N/C} \) at 24 cm from its center. What is the charge?

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

\( 6 \times 10^3 = 9 \times 10^9 \times \frac{q}{(0.24)^2} \).

\( q = \frac{6 \times 10^3 \times 0.0576}{9 \times 10^9} = 3.84 \times 10^{-8} \, \text{C} \).

\( 3.5 \times 10^{-8} \, \text{C} \)
\( 3.7 \times 10^{-8} \, \text{C} \)
\( 3.8 \times 10^{-8} \, \text{C} \)
\( 3.84 \times 10^{-8} \, \text{C} \)
4

A thin spherical shell of radius 16 cm has a charge of \( 11 \, \mu\text{C} \). What is the electric field at a point 20 cm from the center?

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

\( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( q = 11 \times 10^{-6} \, \text{C} \), \( r = 0.2 \, \text{m} \).

\( E = 9 \times 10^9 \times \frac{11 \times 10^{-6}}{(0.2)^2} = 9 \times 10^9 \times \frac{11 \times 10^{-6}}{0.04} = 2.475 \times 10^6 \, \text{N/C} \).

\( 2.475 \times 10^6 \, \text{N/C} \)
\( 2.5 \times 10^6 \, \text{N/C} \)
\( 2.7 \times 10^6 \, \text{N/C} \)
\( 2.9 \times 10^6 \, \text{N/C} \)
1

What property of electric charges explains why two objects with identical charges repel each other?

The polarity of charge (positive or negative) determines interaction: like charges repel due to the repulsive force described by Coulomb’s law, where the force direction depends on the sign of the charges, causing repulsion for identical signs.

Polarity
Quantization
Additivity
Conservation
1

A thin spherical shell of radius 9 cm has \( q = 6 \, \mu\text{C} \). What is the electric field at 12 cm from the center?

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

\( E = 9 \times 10^9 \times \frac{6 \times 10^{-6}}{(0.12)^2} = 9 \times 10^9 \times \frac{6 \times 10^{-6}}{0.0144} = 3.75 \times 10^6 \, \text{N/C} \).

\( 3.5 \times 10^6 \, \text{N/C} \)
\( 3.6 \times 10^6 \, \text{N/C} \)
\( 3.7 \times 10^6 \, \text{N/C} \)
\( 3.75 \times 10^6 \, \text{N/C} \)
4

A net flux of \( 9.04 \times 10^4 \, \text{Nm}^2/\text{C} \) passes through a closed surface. What is the charge enclosed?

\( \phi = \frac{q}{\varepsilon_0} \).

\( q = \phi \varepsilon_0 = 9.04 \times 10^4 \times 8.854 \times 10^{-12} = 8 \times 10^{-7} \, \text{C} = 0.8 \, \mu\text{C} \).

\( 0.6 \, \mu\text{C} \)
\( 0.7 \, \mu\text{C} \)
\( 0.8 \, \mu\text{C} \)
\( 0.9 \, \mu\text{C} \)
3

A conducting sphere of radius 25 cm has an electric field of \( 5 \times 10^3 \, \text{N/C} \) at 50 cm from its center. What is the charge?

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

\( 5 \times 10^3 = 9 \times 10^9 \times \frac{q}{(0.5)^2} \).

\( q = \frac{5 \times 10^3 \times 0.25}{9 \times 10^9} = 1.389 \times 10^{-7} \, \text{C} \).

\( 1.2 \times 10^{-7} \, \text{C} \)
\( 1.3 \times 10^{-7} \, \text{C} \)
\( 1.35 \times 10^{-7} \, \text{C} \)
\( 1.39 \times 10^{-7} \, \text{C} \)
4

What characteristic of an electric field ensures that work done in moving a charge along a closed path is zero in electrostatics?

The conservative nature of the electrostatic field means the work done depends only on the potential difference between points, not the path. For a closed path, the start and end points are the same, so the net work is zero.

Charge density
Conservative nature
Field superposition
Charge symmetry
2

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