Electrostatic Potential and Capacitance — Practice

Question 1

Calculate the potential at a point 20 cm from a charge of 4 x 10^-7 C.

Accepted Answer

Substitute into V = kq/r, remembering the potential falls off as 1/r rather than 1/r^2. Answer: V = 1.8 x 10^4 V.

Question 2

Two capacitors of 4 pF and 12 pF are connected in series. Find the equivalent capacitance.

Accepted Answer

In series the reciprocals add, so 1/C = 1/4 + 1/12 = 1/3, and the result is smaller than either capacitor. Answer: C = 3 pF.

Question 3

Four charges of 2 microcoulomb each are placed at the corners of a square of side 20 cm. Find the potential at the centre.

Accepted Answer

Each corner is a distance of half the diagonal from the centre, which is 0.1414 m, and the four scalar potentials simply add. Answer: V = 5.09 x 10^5 V.

Question 4

Two charges of 3 microcoulomb and -5 microcoulomb are 40 cm apart. Find the potential energy of the system.

Accepted Answer

Use U = kq1q2/r, carrying the signs through so the unlike charges give a negative energy. Answer: U = -0.3375 J.

Question 5

A 20 pF capacitor is charged to 100 V. Find the charge stored and the energy stored.

Accepted Answer

Apply Q = CV and then U = half C V squared, remembering to square the voltage. Answer: Q = 2 x 10^-9 C; U = 1 x 10^-7 J.

Question 6

A parallel plate capacitor of capacitance 10 pF is charged by a 200 V battery, which is then disconnected. A dielectric of constant 5 is inserted. Find the new voltage and the new energy.

Accepted Answer

With the battery disconnected the charge is fixed, so C rises fivefold and V falls to one fifth, and the energy Q squared over 2C falls by the same factor. Answer: V = 40 V; U = 4 x 10^-8 J, one fifth of the original 2 x 10^-7 J.

Question 7

Three charges of 1 microcoulomb each sit at the vertices of an equilateral triangle of side 10 cm. Find the potential energy of the system.

Accepted Answer

There are three pairs, not three charges, and each pair contributes the same amount since all charges and separations are equal. Answer: U = 3 x 9 x 10^9 x 10^-12/0.10 = 0.27 J.

Question 8

A 400 pF capacitor charged to 100 V is disconnected and joined to an uncharged 400 pF capacitor. Find the final voltage and the energy lost.

Accepted Answer

Charge is conserved while the total capacitance doubles, so the voltage halves, and since the capacitors are identical exactly half the energy is dissipated. Answer: V = 50 V; energy lost = 1 x 10^-6 J.

Practice — Electrostatic Potential and Capacitance

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