Coulombs Law and Electric Field

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Q1. What is the capacitance of a capacitor that stores 0.01 J of energy at a voltage of 10 V?

Solution:

Using U = 1/2 C V^2, rearranging gives C = 2U/V^2 = 2(0.01 J)/(10 V)^2 = 0.1 F.

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Q2. A capacitor of 10 µF is charged to a voltage of 5 V. What is the energy stored in the capacitor?

Solution:

Energy stored, U = 0.5 * C * V² = 0.5 * 10 x 10^-6 F * (5 V)² = 0.125 mJ.

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Q3. If the potential difference across a capacitor is doubled, what happens to the stored energy?

Solution:

Energy stored in a capacitor is given by U = 1/2 C V^2. If V is doubled, U becomes 1/2 C (2V)^2 = 2CV^2, which is quadrupled.

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Q4. What is the capacitance of a capacitor that stores 20 µC of charge at a potential difference of 5 V?

Solution:

C = Q/V = 20 x 10^-6 C / 5 V = 4 µF.

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Q5. In a parallel circuit with a 12 V battery and two resistors of 4 Ω and 6 Ω, what is the total current supplied by the battery?

Solution:

Total resistance, R_eq = 1/(1/4 + 1/6) = 2.4 Ω. Total current, I = V/R_eq = 12 V / 2.4 Ω = 5 A.

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Q6. What is the equivalent resistance of three resistors of 4 Ω, 6 Ω, and 12 Ω in series?

Solution:

In series, R_eq = R1 + R2 + R3 = 4 Ω + 6 Ω + 12 Ω = 22 Ω.

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Q7. If the potential difference across a capacitor is 12 V and its capacitance is 3 µF, what is the charge stored in the capacitor?

Solution:

Q = C * V = 3 x 10^-6 F * 12 V = 36 µC.

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Q8. What is the electric field due to a point charge of -4 µC at a distance of 0.25 m?

Solution:

E = k * |q| / r^2 = (8.99 x 10^9 N m²/C²) * (4 x 10^-6 C) / (0.25 m)^2 = -5760 N/C.

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Q9. If two capacitors of 4 µF and 6 µF are connected in series, what is the total capacitance?

Solution:

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

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Q10. If the electric potential at a point is 100 V and the charge at that point is 2 µC, what is the electric potential energy?

Solution:

Potential energy U = V * q = 100 V * 2 x 10^-6 C = 0.0002 J = 0.2 J.