Q. A charge of +1 μC is placed in an electric field of strength 1000 N/C. What is the force experienced by the charge? (2023)
A.
0.001 N
B.
0.01 N
C.
1 N
D.
10 N
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Solution
Force F = qE = (1 × 10^-6 C)(1000 N/C) = 0.001 N.
Correct Answer: B — 0.01 N
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Q. A charge of +10 μC is placed in an electric field of 2000 N/C. What is the force experienced by the charge? (2023)
A.
20 N
B.
200 N
C.
2 N
D.
0.2 N
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Solution
Force F = qE = (10 × 10^-6 C) * (2000 N/C) = 0.02 N = 20 N.
Correct Answer: A — 20 N
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Q. A charge of +10 µC is placed in an electric field of strength 2000 N/C. What is the force acting on the charge? (2023)
A.
20 N
B.
200 N
C.
2 N
D.
0.2 N
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Solution
Force F = qE = (10 × 10^-6 C) * (2000 N/C) = 0.02 N = 20 N.
Correct Answer: A — 20 N
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Q. A charge of +5 µC is placed in an electric field of 300 N/C. What is the work done in moving the charge 0.2 m in the direction of the field? (2019)
A.
30 J
B.
12 J
C.
6 J
D.
3 J
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Solution
Work done W = F * d = (E * q) * d = (300 N/C * 5 × 10^-6 C) * 0.2 m = 0.3 J.
Correct Answer: C — 6 J
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Q. A charge of -2 µC is placed in an electric field of 500 N/C. What is the direction of the force acting on the charge? (2023)
A.
Towards the positive charge
B.
Away from the positive charge
C.
No force
D.
Depends on the field strength
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Solution
The force on a negative charge is in the direction opposite to the electric field, hence towards the positive charge.
Correct Answer: A — Towards the positive charge
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Q. If a charge of +1 µC is placed in an electric field of 1000 N/C, what is the force experienced by the charge? (2019)
A.
0.001 N
B.
1 N
C.
10 N
D.
0.1 N
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Solution
Force F = qE = (1 × 10^-6 C) * (1000 N/C) = 0.001 N.
Correct Answer: A — 0.001 N
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Q. If a charge of +1 µC is placed in an electric field of 2000 N/C, what is the force acting on the charge? (2021)
A.
0.002 N
B.
0.2 N
C.
2 N
D.
20 N
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Solution
Force F = qE = (1 × 10^-6 C) * (2000 N/C) = 0.002 N.
Correct Answer: B — 0.2 N
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Q. If a charge of +10 μC is placed in an electric field of strength 200 N/C, what is the force experienced by the charge? (2019)
A.
200 N
B.
2 N
C.
20 N
D.
0.5 N
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Solution
Force F = qE = (10 × 10^-6 C) * (200 N/C) = 2 N.
Correct Answer: C — 20 N
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Q. If the charge on a capacitor is doubled while keeping the voltage constant, what happens to the capacitance? (2022)
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
Capacitance C = Q/V. If Q is doubled and V remains constant, C must also double.
Correct Answer: A — It doubles
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Q. If the charge on a capacitor is doubled, what happens to the electric field between the plates? (2022)
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
The electric field E between the plates of a capacitor is directly proportional to the charge. If the charge is doubled, the electric field also doubles.
Correct Answer: A — It doubles
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Q. If two charges of +2 µC and -2 µC are placed 1 m apart, what is the net electric field at the midpoint? (2023)
A.
0 N/C
B.
4 N/C
C.
2 N/C
D.
8 N/C
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Solution
The electric fields due to both charges at the midpoint cancel each other out, resulting in a net electric field of 0 N/C.
Correct Answer: A — 0 N/C
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Q. The electric field due to a dipole at a point along the axial line is given by which formula? (2023)
A.
k * p / r²
B.
k * p / r³
C.
k * p / r
D.
k * p / r⁴
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Solution
The electric field E due to a dipole at a point along the axial line is given by E = (1/4πε₀) * (2p) / r³.
Correct Answer: B — k * p / r³
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Q. Two identical charges are placed 1 m apart. If the distance is reduced to 0.5 m, how does the force between them change? (2022)
A.
It doubles
B.
It remains the same
C.
It quadruples
D.
It halves
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Solution
According to Coulomb's law, F ∝ 1/r². Reducing the distance to half increases the force by a factor of 4.
Correct Answer: C — It quadruples
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Q. Two identical charges are placed 1 m apart. If the magnitude of each charge is doubled, what happens to the force between them? (2022)
A.
It remains the same
B.
It doubles
C.
It quadruples
D.
It halves
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Solution
According to Coulomb's law, F ∝ q1 * q2. If both charges are doubled, the force becomes 4 times the original force.
Correct Answer: C — It quadruples
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Q. What happens to the electric field when the charge is doubled while keeping the distance constant? (2023)
A.
It doubles
B.
It halves
C.
It quadruples
D.
It remains the same
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Solution
The electric field E is directly proportional to the charge, so if the charge is doubled, the electric field also doubles.
Correct Answer: A — It doubles
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Q. What is the electric field at a distance of 1 m from a charge of +10 µC? (2019)
A.
9000 N/C
B.
10000 N/C
C.
11000 N/C
D.
12000 N/C
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Solution
E = k * |q| / r² = (9 × 10^9 N m²/C²) * (10 × 10^-6 C) / (1 m)² = 9000 N/C.
Correct Answer: B — 10000 N/C
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Q. What is the electric field at a distance of 1 m from a charge of -4 μC? (2019)
A.
-3600 N/C
B.
-900 N/C
C.
900 N/C
D.
3600 N/C
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Solution
E = k * |q| / r² = (9 × 10^9 N m²/C²) * (4 × 10^-6 C) / (1 m)² = 3600 N/C, directed towards the charge.
Correct Answer: A — -3600 N/C
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Q. What is the electric field at a point due to a uniformly charged infinite plane sheet? (2020)
A.
0
B.
σ/2ε₀
C.
σ/ε₀
D.
σ/4ε₀
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Solution
The electric field due to an infinite plane sheet is E = σ/2ε₀ on both sides, but the total field is σ/ε₀.
Correct Answer: C — σ/ε₀
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Q. What is the electric field due to a point charge of +10 µC at a distance of 0.1 m? (2023)
A.
9000 N/C
B.
900 N/C
C.
90 N/C
D.
9 N/C
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Solution
Electric field E = k * |q| / r² = (9 × 10^9 N m²/C²) * (10 × 10^-6 C) / (0.1 m)² = 9000 N/C.
Correct Answer: A — 9000 N/C
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Q. What is the electric field due to a point charge of +5 µC at a distance of 0.2 m? (2022)
A.
11250 N/C
B.
5625 N/C
C.
22500 N/C
D.
4500 N/C
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Solution
Using E = k * |q| / r² = (9 × 10^9 N m²/C²) * (5 × 10^-6 C) / (0.2 m)² = 11250 N/C.
Correct Answer: A — 11250 N/C
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Q. What is the potential difference between two points in an electric field of 400 N/C separated by 0.5 m? (2022)
A.
200 V
B.
100 V
C.
50 V
D.
80 V
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Solution
Potential difference V = E * d = 400 N/C * 0.5 m = 200 V.
Correct Answer: A — 200 V
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Q. What is the potential energy of a system of two charges +3 µC and +4 µC separated by 0.3 m? (2023)
A.
-0.12 J
B.
0.12 J
C.
0.36 J
D.
0.24 J
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Solution
Potential energy U = k * q1 * q2 / r = (9 × 10^9 N m²/C²) * (3 × 10^-6 C) * (4 × 10^-6 C) / 0.3 m = 0.36 J.
Correct Answer: C — 0.36 J
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Q. What is the potential energy of a system of two charges +3 μC and +4 μC separated by 0.2 m? (2023)
A.
-54 J
B.
54 J
C.
0 J
D.
27 J
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Solution
Potential energy U = k * q1 * q2 / r = (9 × 10^9 N m²/C²) * (3 × 10^-6 C) * (4 × 10^-6 C) / 0.2 m = 54 J.
Correct Answer: B — 54 J
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Q. What is the potential energy of two charges +3 μC and +4 μC separated by 0.2 m? (2023)
A.
-54 J
B.
54 J
C.
0.54 J
D.
5.4 J
Show solution
Solution
Potential energy U = k * q1 * q2 / r = (9 × 10^9 N m²/C²) * (3 × 10^-6 C) * (4 × 10^-6 C) / 0.2 m = 54 J.
Correct Answer: B — 54 J
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Q. Which of the following is a property of electric field lines? (2023)
A.
They can be broken
B.
They can be created
C.
They can be destroyed
D.
They cannot intersect
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Solution
Electric field lines cannot intersect because it would imply two different directions of the electric field at a point.
Correct Answer: D — They cannot intersect
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