Coulombs Law and Electric Field - Capacitance and Dielectrics

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Q1. What is the electric potential at a point 1 m away from a charge of +3 µC?

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.

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Q2. What is the force between two point charges of +2 µC and -3 µC separated by a distance of 0.5 m?

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.

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Q3. What is the equivalent resistance of two resistors, 4 Ω and 6 Ω, connected in parallel?

Solution:

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

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Q4. What is the force between two point charges of +3 µC and -2 µC separated by a distance of 0.5 m?

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.

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Q5. What is the force on a charge of +1 µC placed in an electric field of 500 N/C?

Solution:

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

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Q6. 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?

Solution:

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

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Q7. If a resistor of 10 Ω is connected in series with a 20 Ω resistor, what is the total resistance?

Solution:

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

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Q8. If a capacitor of 10 µF is charged to 20 V, what is the energy stored in the capacitor?

Solution:

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

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Q9. In a circuit with a 12 V battery and a 4 Ω resistor, what is the current flowing through the circuit?

Solution:

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

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Q10. What is the force on a charge of +1 µC placed in an electric field of 1000 N/C?

Solution:

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