Q. A capacitor has a capacitance of 4μF and is charged to 12V. What is the charge on the capacitor?
A.48μC
B.12μC
C.4μC
D.3μC
Solution
Charge (Q) on a capacitor is given by Q = C * V. Thus, Q = 4μF * 12V = 48μC.
Correct Answer: A — 48μC
Q. A capacitor is charged to a potential difference of 12 V. If the capacitance is 4 µF, what is the charge stored in the capacitor?
A.12 µC
B.24 µC
C.48 µC
D.36 µC
Solution
Charge Q is given by Q = CV. Here, Q = 4 µF * 12 V = 48 µC.
Correct Answer: B — 24 µC
Q. A capacitor is charged to a potential difference of V. What is the energy stored in the capacitor?
A.1/2 CV²
B.CV
C.V²/C
D.1/2 QV
Solution
The energy (U) stored in a capacitor is given by U = 1/2 CV², where C is the capacitance and V is the potential difference.
Correct Answer: A — 1/2 CV²
Q. A capacitor is charged to a potential of 12 V. If the capacitance is 3 µF, what is the energy stored in the capacitor?
A.0.18 mJ
B.0.36 mJ
C.0.54 mJ
D.0.72 mJ
Solution
Energy stored in a capacitor is given by U = 1/2 CV² = 1/2 * 3 x 10^-6 F * (12 V)² = 0.36 mJ.
Correct Answer: B — 0.36 mJ
Q. A capacitor is charged to a potential of V. If the charge on the capacitor is doubled, what will be the new potential?
A.V
B.2V
C.V/2
D.4V
Solution
The potential V across a capacitor is directly proportional to the charge. If the charge is doubled, the potential also doubles.
Correct Answer: B — 2V
Q. A capacitor is charged to a voltage V and then disconnected from the battery. If the distance between the plates is increased, what happens to the charge?
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
When a capacitor is disconnected from the battery, the charge remains constant. Increasing the distance decreases capacitance but does not affect the charge.
Correct Answer: C — Remains the same
Q. A capacitor is charged to a voltage V and then disconnected from the battery. If the distance between the plates is doubled, what happens to the voltage across the capacitor?
A.It doubles
B.It halves
C.It remains the same
D.It quadruples
Solution
When the distance is doubled, the capacitance decreases, leading to an increase in voltage since Q = CV is constant.
Correct Answer: A — It doubles
Q. A capacitor is charged to a voltage V and then disconnected from the battery. What happens to the charge on the capacitor if the distance between the plates is increased?
A.Charge increases
B.Charge decreases
C.Charge remains the same
D.Charge becomes zero
Solution
When a capacitor is disconnected from the battery, the charge remains constant. Increasing the distance decreases capacitance but does not change the charge.
Correct Answer: C — Charge remains the same
Q. A capacitor is charged to a voltage V and then disconnected from the battery. What happens to the charge on the capacitor if the voltage is doubled?
A.Charge doubles
B.Charge halves
C.Charge remains the same
D.Charge quadruples
Solution
The charge on a capacitor is given by Q = C * V. If the voltage is doubled, the charge also doubles, assuming capacitance remains constant.
Correct Answer: A — Charge doubles
Q. A capacitor is charged to a voltage V and then the voltage is halved. What happens to the energy stored in the capacitor?
A.It doubles
B.It halves
C.It remains the same
D.It becomes zero
Solution
The energy stored in a capacitor is proportional to the square of the voltage (U = 1/2 CV²). If the voltage is halved, the energy becomes U/4.
Correct Answer: B — It halves
Q. A capacitor of capacitance 10μF is charged to a potential difference of 100V. What is the energy stored in the capacitor?
A.0.05 J
B.0.1 J
C.0.2 J
D.0.3 J
Solution
Energy stored, U = 1/2 * C * V^2 = 1/2 * 10 × 10^-6 * (100)^2 = 0.05 J.
Correct Answer: B — 0.1 J
Q. A capacitor of capacitance 10μF is charged to a potential of 100V. What is the energy stored in the capacitor?
A.0.05 J
B.0.1 J
C.0.2 J
D.0.01 J
Solution
Energy stored in a capacitor is given by U = 1/2 CV² = 1/2 * 10 × 10^-6 F * (100 V)² = 0.05 J.
Correct Answer: B — 0.1 J
Q. A capacitor of capacitance 5μF is charged to a potential of 10V. What is the energy stored in the capacitor?
A.0.25 mJ
B.0.5 mJ
C.1 mJ
D.2.5 mJ
Solution
Energy stored U = 1/2 * C * V² = 1/2 * 5 × 10^-6 F * (10 V)² = 0.5 mJ.
Correct Answer: B — 0.5 mJ
Q. A capacitor of capacitance C is charged to a voltage V and then connected in parallel with another uncharged capacitor of capacitance C. What is the final voltage across the capacitors?
A.V/2
B.V
C.2V
D.0
Solution
When connected in parallel, the total charge is conserved. The final voltage across both capacitors is V/2.
Correct Answer: A — V/2
Q. A capacitor of capacitance C is connected to a battery of voltage V. If the battery is removed and the capacitor is connected to another capacitor of capacitance 2C, what is the final voltage across the combination?
A.V/3
B.V/2
C.V
D.2V
Solution
When the charged capacitor C is connected to an uncharged capacitor 2C, the final voltage is V_final = Q_total / C_eq = V/(1 + 1/2) = V/3.
Correct Answer: B — V/2
Q. A charge of +3μC is placed at the origin. What is the electric potential at a point 0.5m away?
A.5400 V
B.1800 V
C.7200 V
D.3600 V
Solution
V = k * q / r = (9 × 10^9) * (3 × 10^-6) / 0.5 = 54000 V.
Correct Answer: D — 3600 V
Q. A charge of +3μC is placed at the origin. What is the potential at a point 0.3m away?
A.9000 V
B.3000 V
C.10000 V
D.15000 V
Solution
V = k * q / r = (9 × 10^9 N m²/C²) * (3 × 10^-6 C) / 0.3 m = 9000 V.
Correct Answer: A — 9000 V
Q. A charge of +3μC is placed in a uniform electric field of strength 1500 N/C. What is the work done in moving the charge 0.2 m in the direction of the field?
A.90 J
B.60 J
C.30 J
D.45 J
Solution
Work done W = F * d = (E * q) * d = (1500 N/C * 3 × 10^-6 C) * 0.2 m = 0.0009 J = 90 J.
Correct Answer: B — 60 J
Q. A charge of 5 μC is placed in an electric field of 2000 N/C. What is the potential energy of the charge?
A.10 mJ
B.1 mJ
C.0.5 mJ
D.2 mJ
Solution
Potential energy (U) = Charge × Electric Field × Distance. Assuming distance = 1 m, U = 5 μC × 2000 N/C = 10 mJ.
Correct Answer: A — 10 mJ
Q. A charged capacitor has a potential difference of 12 V across its plates. If the capacitance is 4 µF, what is the charge stored in the capacitor?
A.48 µC
B.12 µC
C.3 µC
D.24 µC
Solution
Charge Q = C × V = 4 µF × 12 V = 48 µC.
Correct Answer: A — 48 µC
Q. A charged particle moves from a point of higher electric potential to a point of lower electric potential. What happens to its kinetic energy?
A.Increases
B.Decreases
C.Remains constant
D.Cannot be determined
Solution
As the charged particle moves to a lower potential, it loses potential energy, which is converted into kinetic energy, thus increasing its kinetic energy.
Correct Answer: A — Increases
Q. A charged particle moves from a region of high potential to low potential. What happens to its kinetic energy?
A.It increases
B.It decreases
C.It remains constant
D.It becomes zero
Solution
As the charged particle moves from high potential to low potential, it loses potential energy, which is converted into kinetic energy, thus its kinetic energy increases.
Correct Answer: A — It increases
Q. A charged sphere has a radius R and a total charge Q. What is the electric potential at a point outside the sphere at a distance r from the center (r > R)?
A.kQ/R
B.kQ/r
C.kQ/(R+r)
D.0
Solution
For a charged sphere, the electric potential outside the sphere behaves as if all the charge were concentrated at the center, so V = kQ/r.
Correct Answer: B — kQ/r
Q. A cube of side length a has a charge Q at one of its corners. What is the total electric flux through the cube?
A.Q/ε₀
B.Q/(6ε₀)
C.Q/(12ε₀)
D.0
Solution
The charge at one corner contributes 1/8 of its flux to the cube. Thus, total flux = (1/8)Q/ε₀ = Q/(12ε₀).
Correct Answer: C — Q/(12ε₀)
Q. A cylindrical Gaussian surface encloses a charge Q. If the height of the cylinder is doubled while keeping the radius constant, what happens to the electric flux through the curved surface?
A.It doubles
B.It halves
C.It remains the same
D.It becomes zero
Solution
The electric flux through the curved surface is proportional to the charge enclosed, which remains constant, so the flux through the curved surface doubles if the height is doubled.
Correct Answer: A — It doubles
Q. A cylindrical Gaussian surface encloses a long straight wire carrying a current. What is the electric field at a distance r from the wire?
A.0
B.I/(2πε₀r)
C.λ/(2πε₀r)
D.σ/(2ε₀)
Solution
Gauss's law applies to electric fields, not magnetic fields. The electric field around a current-carrying wire is not defined by Gauss's law.
Correct Answer: A — 0
Q. A cylindrical Gaussian surface encloses a long straight wire carrying a current. What is the electric field at a point outside the cylinder?
A.Zero
B.Directly proportional to the distance from the wire
C.Inversely proportional to the distance from the wire
D.Constant
Solution
The electric field outside the cylindrical surface is directly proportional to the distance from the wire.
Correct Answer: B — Directly proportional to the distance from the wire
Q. A cylindrical Gaussian surface of length L and radius R encloses a charge Q. What is the electric field E at a distance R from the axis of the cylinder?
A.Q/(2πε₀R)
B.Q/(4πε₀R²)
C.Q/(ε₀L)
D.0
Solution
Using Gauss's law, the electric field E at a distance R from the axis of a long charged cylinder is E = Q/(2πε₀L) for points outside the cylinder.
Correct Answer: A — Q/(2πε₀R)
Q. A dipole consists of two charges +q and -q separated by a distance d. What is the expression for the dipole moment?
A.qd
B.q/d
C.q^2d
D.q/d^2
Solution
The dipole moment p is defined as p = q * d.
Correct Answer: A — qd
Q. A dipole consists of two equal and opposite charges separated by a distance of 0.1m. What is the dipole moment if each charge is 1μC?
A.1 × 10^-7 C m
B.1 × 10^-6 C m
C.1 × 10^-5 C m
D.1 × 10^-4 C m
Solution
Dipole moment p = q * d = (1 × 10^-6 C) * (0.1 m) = 1 × 10^-7 C m.
