Coulombs Law and Electric Field

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Q1. Two charges, +4 µC and -4 µC, are placed 1 m apart. What is the net electric field at the midpoint between the charges?

Solution:

The electric fields due to both charges at the midpoint are equal in magnitude and opposite in direction, resulting in a net electric field of 0 N/C.

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Q2. What is the electric potential at a point 1 m away from a +1 C charge?

Solution:

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

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Q3. 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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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. 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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Q6. What is the potential difference across a capacitor of 5 µF charged with 0.01 C?

Solution:

V = Q / C = 0.01 C / 5 x 10^-6 F = 2000 V.

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Q7. 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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Q8. In a parallel circuit with a total voltage of 12 V, what is the voltage across each resistor?

Solution:

In a parallel circuit, the voltage across each resistor is equal to the total voltage, so it is 12 V.

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Q9. 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.

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Q10. 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.