Electrostatics
Q. A charge of +4 µC is placed in a uniform electric field of strength 500 N/C. What is the work done in moving the charge 0.2 m in the direction of the field?
A.
400 J
B.
200 J
C.
100 J
D.
50 J
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Solution
W = F * d = (E * q) * d = (500 N/C * 4 × 10^-6 C) * 0.2 m = 0.4 J.
Correct Answer: B — 200 J
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Q. A parallel plate capacitor has a capacitance of 5 µF. If the potential difference across it is increased from 10 V to 20 V, what is the change in stored energy?
A.
25 µJ
B.
50 µJ
C.
75 µJ
D.
100 µJ
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Solution
Energy U = 1/2 C V^2. Initial U = 1/2 * 5 × 10^-6 * 10^2 = 0.025 J; Final U = 1/2 * 5 × 10^-6 * 20^2 = 0.1 J. Change = 0.1 - 0.025 = 0.075 J = 75 µJ.
Correct Answer: B — 50 µJ
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Q. If a charge of +2μC is placed in an electric field of 1000 N/C, what is the force experienced by the charge? (2023)
A.
2 N
B.
0.002 N
C.
2000 N
D.
0.2 N
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Solution
The force (F) on a charge in an electric field is given by F = qE. Here, F = 2μC * 1000 N/C = 2 N.
Correct Answer: A — 2 N
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Q. If the charge on a capacitor is doubled, what happens to the energy stored? (2019)
A.
It remains the same
B.
It doubles
C.
It quadruples
D.
It halves
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Solution
The energy stored in a capacitor is proportional to the square of the charge (U = Q^2/(2C)). Doubling the charge increases the energy by a factor of 4.
Correct Answer: C — It quadruples
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Q. If the electric field at a point is 1000 N/C and the charge at that point is 5 µC, what is the force experienced by the charge?
A.
0.005 N
B.
0.05 N
C.
0.5 N
D.
5 N
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Solution
F = E * q = 1000 N/C * 5 × 10^-6 C = 0.005 N.
Correct Answer: B — 0.05 N
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Q. If two charges +Q and -Q are placed at a distance 'd' apart, what is the electric field at the midpoint? (2020)
A.
0
B.
kQ/d^2
C.
kQ/(2d^2)
D.
kQ/d
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Solution
At the midpoint, the electric fields due to +Q and -Q cancel each other out, resulting in a net electric field of 0.
Correct Answer: A — 0
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Q. If two identical capacitors of 4 µF each are connected in series, what is the equivalent capacitance?
A.
2 µF
B.
4 µF
C.
6 µF
D.
8 µF
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Solution
C_eq = 1 / (1/C1 + 1/C2) = 1 / (1/4 + 1/4) = 2 µF.
Correct Answer: A — 2 µF
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Q. What is the capacitance of a capacitor that stores 10 µC of charge at a potential difference of 5 V?
A.
2 µF
B.
5 µF
C.
10 µF
D.
20 µF
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Solution
C = Q/V = 10 × 10^-6 C / 5 V = 2 × 10^-6 F = 2 µF.
Correct Answer: A — 2 µF
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Q. What is the effect of a dielectric material on the capacitance of a capacitor? (2023)
A.
It decreases the capacitance
B.
It increases the capacitance
C.
It has no effect
D.
It makes the capacitor discharge
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Solution
Inserting a dielectric material increases the capacitance of a capacitor by reducing the electric field within it.
Correct Answer: B — It increases the capacitance
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Q. What is the electric field due to a point charge at a distance 'r' from it? (2021)
A.
k * q / r^2
B.
k * q / r
C.
k * q * r
D.
k * q * r^2
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Solution
The electric field (E) due to a point charge (q) at a distance (r) is given by E = k * q / r^2, where k is Coulomb's constant.
Correct Answer: A — k * q / r^2
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Q. What is the electric field due to a point charge of +3 µC at a distance of 0.3 m?
A.
30 N/C
B.
90 N/C
C.
300 N/C
D.
900 N/C
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Solution
E = k * |q| / r^2 = (9 × 10^9) * (3 × 10^-6) / (0.3)^2 = 90 N/C.
Correct Answer: B — 90 N/C
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Q. What is the electric potential at a point 0.4 m away from a charge of +5 µC?
A.
45 V
B.
50 V
C.
55 V
D.
60 V
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Solution
V = k * q / r = (9 × 10^9) * (5 × 10^-6) / 0.4 = 112.5 V.
Correct Answer: B — 50 V
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Q. What is the electric potential at a point due to a charge Q at a distance r? (2021)
A.
kQ/r
B.
kQr
C.
kQ/r^2
D.
0
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Solution
The electric potential (V) at a distance r from a point charge Q is given by V = kQ/r.
Correct Answer: A — kQ/r
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Q. What is the force between two point charges of +2 µC and -3 µC separated by a distance of 0.5 m in vacuum?
A.
-24 N
B.
-12 N
C.
12 N
D.
24 N
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Solution
Using Coulomb's law, F = k * |q1 * q2| / r^2 = (9 × 10^9) * |2 × 10^-6 * -3 × 10^-6| / (0.5)^2 = -24 N.
Correct Answer: A — -24 N
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Q. What is the potential energy of a system of two charges of +1 µC and +2 µC separated by 0.1 m?
A.
180 J
B.
90 J
C.
45 J
D.
36 J
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Solution
U = k * (q1 * q2) / r = (9 × 10^9) * (1 × 10^-6 * 2 × 10^-6) / 0.1 = 36 J.
Correct Answer: D — 36 J
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