Electric Charges and Fields Chapter-Wise Test 5

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Two charges \( +4 \, \mu\text{C} \) and \( -6 \, \mu\text{C} \) are 30 cm apart. What is the electric field at a point 15 cm from \( +4 \, \mu\text{C} \) along the line joining them?

Distance from \( -6 \, \mu\text{C} = 15 \, \text{cm} = 0.15 \, \text{m} \).

\( E_1 = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{(0.15)^2} = 1.6 \times 10^6 \, \text{N/C} \) (away).

\( E_2 = 9 \times 10^9 \times \frac{6 \times 10^{-6}}{(0.15)^2} = 2.4 \times 10^6 \, \text{N/C} \) (towards).

Net \( E = 2.4 \times 10^6 - 1.6 \times 10^6 = 8 \times 10^5 \, \text{N/C} \) (towards \( -6 \, \mu\text{C} \)).

\( 7.5 \times 10^5 \, \text{N/C} \)
\( 8.0 \times 10^5 \, \text{N/C} \)
\( 8.5 \times 10^5 \, \text{N/C} \)
\( 9.0 \times 10^5 \, \text{N/C} \)
2

What explains why an electric dipole aligns with an external field when free to rotate?

The torque on a dipole (\( \tau = pE \sin \theta \)) acts to minimize potential energy, aligning the dipole with the field (where \( \theta = 0 \), torque = 0). This rotational force drives the dipole to a stable orientation parallel to the field.

Charge quantization
Field symmetry
Charge separation
Torque
4

Three charges \( +4 \, \mu\text{C}, -2 \, \mu\text{C}, +5 \, \mu\text{C} \) are at the vertices of an equilateral triangle of side 1.2 m. What is the force magnitude on \( +4 \, \mu\text{C} \)?

\( F_1 = 9 \times 10^9 \times \frac{4 \times 2 \times 10^{-12}}{(1.2)^2} = 0.05 \, \text{N} \) (attractive).

\( F_2 = 9 \times 10^9 \times \frac{4 \times 5 \times 10^{-12}}{(1.2)^2} = 0.125 \, \text{N} \) (repulsive).

Angle 60°. Net \( F = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos 60^\circ} = \sqrt{0.05^2 + 0.125^2 + 0.00625} = 0.144 \, \text{N} \).

0.12 N
0.13 N
0.144 N
0.15 N
3

A charge \( 6 \, \mu\text{C} \) is at the center of a cube of edge 20 cm. What is the total electric flux through the cube?

Total flux: \( \phi = \frac{q}{\varepsilon_0} \).

\( \phi = \frac{6 \times 10^{-6}}{8.854 \times 10^{-12}} = 6.78 \times 10^5 \, \text{Nm}^2/\text{C} \).

\( 6.0 \times 10^5 \, \text{Nm}^2/\text{C} \)
\( 6.78 \times 10^5 \, \text{Nm}^2/\text{C} \)
\( 7.5 \times 10^5 \, \text{Nm}^2/\text{C} \)
\( 8.0 \times 10^5 \, \text{Nm}^2/\text{C} \)
2

A dipole has charges \( +q = 3 \, \mu\text{C} \) and \( -q = -3 \, \mu\text{C} \) separated by 2 mm. What is the dipole moment?

Dipole moment: \( p = q \times 2a \), where \( 2a = 2 \times 10^{-3} \, \text{m} \), \( q = 3 \times 10^{-6} \, \text{C} \).

\( p = 3 \times 10^{-6} \times 2 \times 10^{-3} = 6 \times 10^{-9} \, \text{C m} \).

\( 3 \times 10^{-9} \, \text{C m} \)
\( 6 \times 10^{-9} \, \text{C m} \)
\( 9 \times 10^{-9} \, \text{C m} \)
\( 1.2 \times 10^{-8} \, \text{C m} \)
2

Two point charges \( 8 \times 10^{-7} \, \text{C} \) and \( -4 \times 10^{-7} \, \text{C} \) are 60 cm apart in vacuum. What is the magnitude of the force between them?

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

\( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( q_1 = 8 \times 10^{-7} \, \text{C} \), \( q_2 = -4 \times 10^{-7} \, \text{C} \), \( r = 0.6 \, \text{m} \).

\( |q_1 q_2| = 8 \times 4 \times 10^{-14} = 32 \times 10^{-14} \, \text{C}^2 \).

\( r^2 = (0.6)^2 = 0.36 \, \text{m}^2 \).

\( F = 9 \times 10^9 \times \frac{32 \times 10^{-14}}{0.36} = 9 \times 10^9 \times 8.89 \times 10^{-13} = 0.008 \, \text{N} \).

0.008 N
0.009 N
0.010 N
0.012 N
1

Why does the torque on an electric dipole in a uniform field depend on the sine of the angle between the dipole and the field?

Torque (\( \tau = pE \sin \theta \)) arises from the couple formed by forces on the dipole’s charges. The perpendicular component of the field to the dipole axis determines the rotational effect, which is maximum at 90° and zero when aligned (sin 0° = 0).

Charge symmetry
Field uniformity
Charge separation
Perpendicular component
4

A dipole with \( p = 6 \times 10^{-9} \, \text{C m} \) is at 60° to a field \( E = 3 \times 10^4 \, \text{N/C} \). What is the torque?

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

\( \tau = 6 \times 10^{-9} \times 3 \times 10^4 \times \sin 60^\circ = 18 \times 10^{-5} \times \frac{\sqrt{3}}{2} = 1.56 \times 10^{-4} \, \text{N m} \).

\( 1.2 \times 10^{-4} \, \text{N m} \)
\( 1.4 \times 10^{-4} \, \text{N m} \)
\( 1.5 \times 10^{-4} \, \text{N m} \)
\( 1.56 \times 10^{-4} \, \text{N m} \)
4

A closed surface has a net flux of \( 4.52 \times 10^5 \, \text{Nm}^2/\text{C} \). What is the charge enclosed?

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

\( q = \phi \varepsilon_0 = 4.52 \times 10^5 \times 8.854 \times 10^{-12} = 4 \times 10^{-6} \, \text{C} = 4 \, \mu\text{C} \).

\( 3.8 \, \mu\text{C} \)
\( 3.9 \, \mu\text{C} \)
\( 4.0 \, \mu\text{C} \)
\( 4.1 \, \mu\text{C} \)
3

A closed surface has a net flux of \( 2.26 \times 10^5 \, \text{Nm}^2/\text{C} \). What is the charge enclosed?

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

\( q = \phi \varepsilon_0 = 2.26 \times 10^5 \times 8.854 \times 10^{-12} = 2 \times 10^{-6} \, \text{C} = 2 \, \mu\text{C} \).

\( 1.8 \, \mu\text{C} \)
\( 1.9 \, \mu\text{C} \)
\( 2.0 \, \mu\text{C} \)
\( 2.1 \, \mu\text{C} \)
3

Two point charges \( 4 \times 10^{-7} \, \text{C} \) and \( -8 \times 10^{-7} \, \text{C} \) are 90 cm apart in vacuum. What is the magnitude of the force between them?

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

\( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( q_1 = 4 \times 10^{-7} \, \text{C} \), \( q_2 = -8 \times 10^{-7} \, \text{C} \), \( r = 0.9 \, \text{m} \).

\( |q_1 q_2| = 4 \times 8 \times 10^{-14} = 32 \times 10^{-14} \, \text{C}^2 \).

\( r^2 = (0.9)^2 = 0.81 \, \text{m}^2 \).

\( F = 9 \times 10^9 \times \frac{32 \times 10^{-14}}{0.81} = 9 \times 10^9 \times 3.951 \times 10^{-13} = 0.00356 \, \text{N} \).

0.00356 N
0.0038 N
0.0040 N
0.0042 N
1

What ensures that the electric field outside a neutral conductor in an external field is altered by induced charges?

Induced charge separation occurs as the external field polarizes the conductor, creating surface charges. These charges generate an additional field that superposes with the external field, modifying the total field outside the conductor.

Charge quantization
Field cancellation
Induced charge separation
Charge symmetry
3

Which property of electric charges is responsible for the fact that charges on a conductor in equilibrium reside only on its surface?

In electrostatic equilibrium, the electric field inside a conductor must be zero. If charges existed inside, they would create a field, causing further movement. Thus, charges redistribute to the surface, where they can maintain zero internal field due to their mobility.

Mobility
Quantization
Additivity
Symmetry
1

A plastic rod gains a charge of \( -1.92 \times 10^{-7} \, \text{C} \) when rubbed. How many electrons were transferred to it?

Negative charge means electrons gained.

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

\( n = \frac{q}{|e|} = \frac{1.92 \times 10^{-7}}{1.6 \times 10^{-19}} = 1.2 \times 10^{12} \).

\( 1.2 \times 10^{12} \)
\( 1.3 \times 10^{12} \)
\( 1.4 \times 10^{12} \)
\( 1.5 \times 10^{12} \)
1

Why does the electric field due to an infinite charged sheet remain constant regardless of its thickness?

For an infinite sheet, Gauss’s law shows the field depends only on the surface charge density (\( \sigma/2\varepsilon_0 \)). Thickness doesn’t affect the planar symmetry or enclosed charge per unit area, keeping the field constant.

Charge quantization
Field symmetry
Charge mobility
Surface charge density
4

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