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

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Q. If a capacitor of 10 µF is connected in series with a 20 µF capacitor, what is the total capacitance?

A. 6.67 µF
B. 30 µF
C. 15 µF
D. 8 µF

Solution

1/C_total = 1/C1 + 1/C2 = 1/10 + 1/20 = 3/20, thus C_total = 20/3 µF = 6.67 µF.

Correct Answer: A — 6.67 µF

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Q. If a dielectric material with a dielectric constant of 4 is placed in a capacitor, how does it affect the capacitance?

A. It halves the capacitance
B. It doubles the capacitance
C. It quadruples the capacitance
D. It does not affect the capacitance

Solution

The capacitance increases by the factor of the dielectric constant, so C' = k * C, where k = 4.

Correct Answer: B — It doubles the capacitance

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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

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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

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Q. In a series circuit with a 12 V battery and two resistors of 3 Ω and 6 Ω, what is the current flowing through the circuit?

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

Solution

Total resistance R_total = R1 + R2 = 3 Ω + 6 Ω = 9 Ω. Current I = V/R_total = 12 V / 9 Ω = 1.33 A, approximately 1 A.

Correct Answer: B — 2 A

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

A. 2 µF
B. 4 µF
C. 6 µF
D. 8 µF

Solution

C = Q/V = 12 x 10^-6 C / 6 V = 2 x 10^-6 F = 2 µF.

Correct Answer: B — 4 µF

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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

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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=2?

A. 5.5 µF
B. 4.4 µF
C. 2.2 µF
D. 1.1 µF

Solution

C = k * ε₀ * (A/d) = 2 * (8.85 x 10^-12 F/m) * (0.01 m² / 0.001 m) = 4.4 µF.

Correct Answer: B — 4.4 µF

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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

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Q. What is the current through a 5 Ω resistor connected to a 15 V battery?

A. 3 A
B. 2 A
C. 1 A
D. 0.5 A

Solution

Using Ohm's law, I = V/R = 15 V / 5 Ω = 3 A.

Correct Answer: B — 2 A

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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

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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

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

A. 9 V
B. 27 V
C. 18 V
D. 36 V

Solution

V = k * q / r = (8.99 x 10^9 N m²/C²) * (3 x 10^-6 C) / (1 m) = 26.97 V, approximately 27 V.

Correct Answer: B — 27 V

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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

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Q. What is the electric potential energy stored in a capacitor of 2 µF charged to 12 V?

A. 0.144 mJ
B. 0.288 mJ
C. 0.576 mJ
D. 0.072 mJ

Solution

U = 1/2 * C * V² = 1/2 * 2 x 10^-6 F * (12 V)² = 0.144 mJ.

Correct Answer: B — 0.288 mJ

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Q. What is the energy stored in a capacitor of 5 µF charged to a voltage of 10 V?

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

Solution

Energy (U) = 0.5 * C * V² = 0.5 * 5 x 10^-6 F * (10 V)² = 0.25 mJ.

Correct Answer: A — 0.25 mJ

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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

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

A. 10 Ω
B. 5 Ω
C. 8 Ω
D. 12 Ω

Solution

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

Correct Answer: A — 10 Ω

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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

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Q. What is the equivalent resistance of two resistors, 4 Ω and 6 Ω, connected in parallel?

A. 2.4 Ω
B. 10 Ω
C. 24 Ω
D. 1.5 Ω

Solution

1/R_total = 1/R1 + 1/R2 = 1/4 + 1/6 = 5/12, thus R_total = 12/5 Ω = 2.4 Ω.

Correct Answer: A — 2.4 Ω

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Q. What is the force between two charges of +1 µC and +1 µC separated by 0.1 m?

A. 0.09 N
B. 0.18 N
C. 0.02 N
D. 0.04 N

Solution

F = k * |q1 * q2| / r² = (8.99 x 10^9 N m²/C²) * (1 x 10^-6 C)² / (0.1 m)² = 0.09 N.

Correct Answer: B — 0.18 N

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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

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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?

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

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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 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

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

A. 1.2 N
B. 0.6 N
C. 0.4 N
D. 0.8 N

Solution

Using Coulomb's law, F = k * |q1 * q2| / r^2 = (8.99 x 10^9 N m²/C²) * |3 x 10^-6 C * -2 x 10^-6 C| / (0.5 m)² = 1.2 N.

Correct Answer: A — 1.2 N

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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

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Q. What is the force on a charge of +1 µC placed in an electric field of 500 N/C?

A. 0.5 N
B. 1 N
C. 2 N
D. 0.2 N

Solution

F = q * E = (1 x 10^-6 C) * (500 N/C) = 0.5 N.

Correct Answer: A — 0.5 N

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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

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

A. 0.05 V
B. 0.5 V
C. 5 V
D. 50 V

Solution

V = Q/C; Q = C * V = 5 µF * 10 V = 50 µC, thus the potential difference is 10 V.

Correct Answer: C — 5 V

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Showing 1 to 30 of 35 (2 Pages)

Coulombs Law and Electric Field - Capacitance and Dielectrics

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