Coulombs Law and Electric Field - Capacitance and Dielectrics

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Q1. What is the current through a 5 Ω resistor connected to a 15 V battery?

Solution:

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

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Q2. What is the total energy stored in a 10 µF capacitor charged to 20 V?

Solution:

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

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Q3. 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?

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.

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Q4. What is the electric potential at a point 1 m away from a charge of +1 µC?

Solution:

V = k * q / r = (8.99 x 10^9 N m²/C²) * (1 x 10^-6 C) / 1 m = 8.99 kV.

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Q5. What is the potential difference across a capacitor of 5 µF charged to 10 V?

Solution:

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

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Q6. If a capacitor of 10 µF is connected in series with a 20 µF capacitor, what is the total capacitance?

Solution:

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

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Q7. What is the electric potential at a point 3 m away from a charge of +1 µC?

Solution:

V = k * q / r = (8.99 x 10^9 N m²/C²) * (1 x 10^-6 C) / 3 m = 3000 V.

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Q8. What is the potential difference across a 5 µF capacitor charged to 10 V?

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.

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Q9. 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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Q10. 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.