Question

Two point charges q₁ and q₂ are kept at a distance of r₁₂ in air. Deduce the expression for the electrostatic potential energy of this system.

Answer 1

The work done to move q₁ from infinity to A is, W₁ = 0 (∵ There is no initial electric field)

The work done to move q₂ from infinity to B is, W₂ = V_1Bq₂

Where V_1B is the electric potential at B due to q₁, it is given by

\(V_{1B} = \frac{1}{4 \pi \epsilon_0 \left(\frac{q_1}{r_{12}}\right)}\)

\(\Rightarrow W_2 = V_{1B} q_2 = \frac{1}{4 \pi \epsilon_0} \left(\frac{q_1}{q_2}\right) q_2\)

The potential energy of this system of charge is equal to total work done to build this configuration. Therefore

\(U= W_1 + W_2 = 0 + \frac{1}{4 \pi \epsilon_0} \left(\frac{q_1 q_2}{r_{12}}\right)\)

\(\Rightarrow U = \frac{1}{4 \pi \epsilon_0} \left(\frac{q_1q_2}{r_{12}}\right)\)