Coulombs Law and Electric Field

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Q1. If two charges of +3 µC and -3 µC are placed 0.2 m apart, what is the magnitude of the force between them?

Solution:

Using Coulomb's law, F = k * |q1 * q2| / r^2 = (8.99 x 10^9 N m²/C²) * |3 x 10^-6 C * -3 x 10^-6 C| / (0.2 m)^2 = 45 N.

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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 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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Q4. 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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Q5. 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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Q6. What is the electric field between two parallel plates separated by 0.1 m with a potential difference of 100 V?

Solution:

E = V/d = 100 V / 0.1 m = 1000 N/C.

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Q7. 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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Q8. What is the electric field strength at a distance of 2 m from a charge of +5 µC?

Solution:

E = k * |q| / r^2 = (8.99 x 10^9 N m²/C²) * |5 x 10^-6 C| / (2 m)^2 = 1.125 N/C.

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Q9. 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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Q10. 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.