Coulombs Law and Electric Field - Exam Prep Insights

Coulombs Law and Electric Field - Capacitance and Dielectrics

Q. If a capacitor of 10 µF is charged to 20 V, what is the energy stored in the capacitor?

A. 0.02 J
B. 0.1 J
C. 0.5 J
D. 0.1 mJ

Solution

Energy stored, U = 0.5 * C * V^2 = 0.5 * 10 x 10^-6 F * (20 V)^2 = 0.02 J.

Correct Answer: B — 0.1 J

Q. If a resistor of 10 Ω is connected in series with a 20 Ω resistor, what is the total resistance?

A. 10 Ω
B. 20 Ω
C. 30 Ω
D. 15 Ω

Solution

R_total = R1 + R2 = 10 Ω + 20 Ω = 30 Ω.

Correct Answer: C — 30 Ω

Q. In a circuit with a 12 V battery and a 4 Ω resistor, what is the current flowing through the circuit?

A. 3 A
B. 2 A
C. 4 A
D. 1 A

Solution

Using Ohm's law, I = V/R = 12 V / 4 Ω = 3 A.

Correct Answer: B — 2 A

Q. What is the capacitance of a capacitor with a charge of 6 µC and a voltage of 12 V?

A. 0.5 µF
B. 0.25 µF
C. 0.75 µF
D. 1 µF

Solution

C = Q/V = 6 x 10^-6 C / 12 V = 0.5 x 10^-6 F = 0.5 µF.

Correct Answer: A — 0.5 µF

Q. What is the capacitance of a parallel plate capacitor with an area of 0.01 m² and a separation of 0.001 m, filled with a dielectric of k=5?

A. 5.5 µF
B. 5.0 µF
C. 4.5 µF
D. 6.0 µF

Solution

C = k * ε0 * (A/d) = 5 * (8.85 x 10^-12 F/m) * (0.01 m² / 0.001 m) = 5.0 µF.

Correct Answer: B — 5.0 µF

Q. What is the electric field strength at a distance of 2 m from a point charge of +5 µC?

A. 1.12 N/C
B. 0.56 N/C
C. 2.25 N/C
D. 0.75 N/C

Solution

E = k * |q| / r^2 = (8.99 x 10^9 N m²/C²) * (5 x 10^-6 C) / (2 m)^2 = 1.12 N/C.

Correct Answer: A — 1.12 N/C

Q. What is the electric potential at a point 1 m away from a charge of +1 µC?

A. 9 kV
B. 1 kV
C. 0.9 kV
D. 0.1 kV

Solution

V = k * q / r = (8.99 x 10^9 N m²/C²) * (1 x 10^-6 C) / 1 m = 8.99 kV.

Correct Answer: A — 9 kV

Q. What is the electric potential at a point 3 m away from a charge of +1 µC?

A. 3000 V
B. 9000 V
C. 300 V
D. 900 V

Solution

V = k * q / r = (8.99 x 10^9 N m²/C²) * (1 x 10^-6 C) / 3 m = 3000 V.

Correct Answer: C — 300 V

Q. What is the equivalent capacitance of two capacitors, 4 µF and 6 µF, connected in series?

A. 2.4 µF
B. 10 µF
C. 1.5 µF
D. 3.6 µF

Solution

1/C_eq = 1/C1 + 1/C2 = 1/4 + 1/6 => C_eq = 2.4 µF.

Correct Answer: A — 2.4 µF

Q. What is the equivalent resistance of three resistors of 2 Ω, 3 Ω, and 5 Ω in series?

A. 10 Ω
B. 8 Ω
C. 5 Ω
D. 3 Ω

Solution

R_eq = R1 + R2 + R3 = 2 Ω + 3 Ω + 5 Ω = 10 Ω.

Correct Answer: A — 10 Ω

Q. What is the force between two charges of +1 µC and +1 µC separated by 1 m?

A. 8.99 N
B. 0.009 N
C. 0.089 N
D. 0.899 N

Solution

F = k * |q1 * q2| / r^2 = (8.99 x 10^9 N m²/C²) * (1 x 10^-6 C)² / (1 m)² = 8.99 N.

Correct Answer: A — 8.99 N

Q. What is the force between two point charges of +2 µC and -3 µC separated by a distance of 0.5 m in a vacuum?

A. -1.08 N
B. -0.72 N
C. 1.08 N
D. 0.72 N

Solution

Using Coulomb's law, F = k * |q1 * q2| / r^2 = (8.99 x 10^9 N m²/C²) * |(2 x 10^-6 C) * (-3 x 10^-6 C)| / (0.5 m)^2 = -1.08 N.

Correct Answer: A — -1.08 N

Q. What is the force between two point charges of +2 µC and -3 µC separated by a distance of 0.5 m?

A. 1.2 N
B. 0.24 N
C. 0.48 N
D. 0.96 N

Solution

Using Coulomb's law, F = k * |q1 * q2| / r^2 = (8.99 x 10^9 N m²/C²) * |(2 x 10^-6 C) * (-3 x 10^-6 C)| / (0.5 m)^2 = 0.24 N.

Correct Answer: B — 0.24 N

Q. What is the force on a charge of +1 µC placed in an electric field of 1000 N/C?

A. 0.001 N
B. 0.1 N
C. 1 N
D. 10 N

Solution

F = q * E = (1 x 10^-6 C) * (1000 N/C) = 0.001 N.

Correct Answer: B — 0.1 N

Q. What is the potential difference across a 5 µF capacitor charged to 10 V?

A. 50 mJ
B. 0.05 J
C. 0.5 J
D. 5 J

Solution

Energy stored in a capacitor, U = 0.5 * C * V^2 = 0.5 * (5 x 10^-6 F) * (10 V)^2 = 0.05 J.

Correct Answer: B — 0.05 J

Q. What is the total capacitance of two capacitors of 4 µF and 6 µF in series?

A. 2.4 µF
B. 10 µF
C. 1.5 µF
D. 3.6 µF

Solution

1/C_eq = 1/C1 + 1/C2 = 1/4 + 1/6 = 5/12, thus C_eq = 12/5 = 2.4 µF.

Correct Answer: A — 2.4 µF

Q. What is the total energy stored in a 10 µF capacitor charged to 20 V?

A. 2 J
B. 0.02 J
C. 0.2 J
D. 0.5 J

Solution

U = 0.5 * C * V^2 = 0.5 * (10 x 10^-6 F) * (20 V)^2 = 0.02 J.

Correct Answer: C — 0.2 J

Q. What is the voltage across a 10 Ω resistor carrying a current of 2 A?

A. 5 V
B. 10 V
C. 20 V
D. 15 V

Solution

Using Ohm's law, V = I * R = 2 A * 10 Ω = 20 V.

Correct Answer: B — 10 V

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