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  4. Electric Charges and Fields Chapter-Wise Test 2

Electric Charges and Fields Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Which characteristic of insulators prevents them from shielding their interior from an external electric field?

Insulators lack free charges that can move to cancel an external field. Unlike conductors, where mobile electrons redistribute, insulators’ fixed charges allow the field to penetrate, as no shielding mechanism exists.

Absence of free charges
High charge density
Field symmetry
Charge quantization
1

A uniform electric field \( E = 4 \times 10^3 \, \text{N/C} \) is along the z-axis. What is the flux through a square of side 15 cm in the xy-plane?

Area vector \( \Delta \mathbf{S} = (0.15)^2 = 0.0225 \, \text{m}^2 \) along z-axis.

Flux: \( \phi = \mathbf{E} \cdot \Delta \mathbf{S} = 4 \times 10^3 \times 0.0225 = 90 \, \text{Nm}^2/\text{C} \).

\( 90 \, \text{Nm}^2/\text{C} \)
\( 120 \, \text{Nm}^2/\text{C} \)
\( 150 \, \text{Nm}^2/\text{C} \)
\( 180 \, \text{Nm}^2/\text{C} \)
1

Two point charges \( 5 \times 10^{-7} \, \text{C} \) and \( -7 \times 10^{-7} \, \text{C} \) are 150 cm apart in vacuum. What is the magnitude of the force between them?

Using Coulomb’s law: \( F = k \frac{|q_1 q_2|}{r^2} \).

\( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( q_1 = 5 \times 10^{-7} \, \text{C} \), \( q_2 = -7 \times 10^{-7} \, \text{C} \), \( r = 1.5 \, \text{m} \).

\( |q_1 q_2| = 5 \times 7 \times 10^{-14} = 35 \times 10^{-14} \, \text{C}^2 \).

\( r^2 = (1.5)^2 = 2.25 \, \text{m}^2 \).

\( F = 9 \times 10^9 \times \frac{35 \times 10^{-14}}{2.25} = 9 \times 10^9 \times 1.556 \times 10^{-13} = 0.0014 \, \text{N} \).

0.0014 N
0.0015 N
0.0016 N
0.0017 N
1

Why can’t the electric field inside a charged insulator be zero, unlike in a conductor?

In insulators, charges are fixed and cannot move to cancel an internal field. If charges are present inside, they generate a field that persists, as there are no free charges to redistribute and neutralize it, unlike in conductors.

Charge quantization
Field symmetry
Lack of free charges
Charge density
3

A thin spherical shell of radius 8 cm has \( q = 4 \, \mu\text{C} \). What is the electric field at 10 cm from the center?

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

\( E = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{(0.1)^2} = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{0.01} = 3.6 \times 10^6 \, \text{N/C} \).

\( 3.0 \times 10^6 \, \text{N/C} \)
\( 3.2 \times 10^6 \, \text{N/C} \)
\( 3.4 \times 10^6 \, \text{N/C} \)
\( 3.6 \times 10^6 \, \text{N/C} \)
4

An infinite line charge has \( E = 5.4 \times 10^5 \, \text{N/C} \) at 6 cm. What is \( \lambda \)?

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

\( 5.4 \times 10^5 = \frac{2 \times 9 \times 10^9 \times \lambda}{0.06} \).

\( \lambda = \frac{5.4 \times 10^5 \times 0.06}{18 \times 10^9} = 1.8 \times 10^{-6} \, \text{C/m} \).

\( 1.6 \times 10^{-6} \, \text{C/m} \)
\( 1.7 \times 10^{-6} \, \text{C/m} \)
\( 1.75 \times 10^{-6} \, \text{C/m} \)
\( 1.8 \times 10^{-6} \, \text{C/m} \)
4

A dipole with \( p = 6 \times 10^{-9} \, \text{C m} \) is at 60° to a field \( E = 7 \times 10^4 \, \text{N/C} \). What is the torque?

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

\( \tau = 6 \times 10^{-9} \times 7 \times 10^4 \times \sin 60^\circ = 42 \times 10^{-5} \times \frac{\sqrt{3}}{2} = 3.64 \times 10^{-4} \, \text{N m} \).

\( 3.4 \times 10^{-4} \, \text{N m} \)
\( 3.5 \times 10^{-4} \, \text{N m} \)
\( 3.6 \times 10^{-4} \, \text{N m} \)
\( 3.64 \times 10^{-4} \, \text{N m} \)
4

What property of electric charges explains why the smallest observable charge is that of an electron or proton?

Quantization of charge means that charge exists in discrete units, with the smallest unit being the charge of an electron or proton (\( e \)). All observable charges are integer multiples of this fundamental unit, reflecting this property.

Conservation
Quantization
Additivity
Polarity
2

A metallic sphere of radius 15 cm has a charge of \( 6 \, \mu\text{C} \). What is the electric field just outside its surface?

For a conductor, \( E = \frac{k q}{r^2} \) at surface (\( r = 0.15 \, \text{m} \)).

\( E = 9 \times 10^9 \times \frac{6 \times 10^{-6}}{(0.15)^2} = 2.4 \times 10^6 \, \text{N/C} \).

\( 1.8 \times 10^6 \, \text{N/C} \)
\( 2.4 \times 10^6 \, \text{N/C} \)
\( 3.0 \times 10^6 \, \text{N/C} \)
\( 3.6 \times 10^6 \, \text{N/C} \)
2

Two charges \( +9 \, \mu\text{C} \) and \( -6 \, \mu\text{C} \) are 45 cm apart. What is the distance from \( +9 \, \mu\text{C} \) where the electric field is zero?

Let \( x \) be distance from \( +9 \, \mu\text{C} \), then \( 0.45 - x \) from \( -6 \, \mu\text{C} \).

\( \frac{9 \times 10^{-6}}{x^2} = \frac{6 \times 10^{-6}}{(0.45 - x)^2} \), \( 9 (0.45 - x)^2 = 6 x^2 \).

\( 1.5 (0.2025 - 0.9 x + x^2) = x^2 \), \( 0.30375 - 1.35 x + 1.5 x^2 = x^2 \).

\( 0.5 x^2 - 1.35 x + 0.30375 = 0 \), \( x = \frac{1.35 \pm \sqrt{1.8225 - 0.6075}}{1} = \frac{1.35 \pm 1.05}{1} \).

\( x = 0.3 \, \text{m} \) (between charges).

25 cm
28 cm
29 cm
30 cm
4

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