Question

Two charges 5 nC and –2 nC are placed at points (5 cm, 0, 0) and (23 cm, 0, 0) in a region of space where there is no other external field. Calculate the electrostatic potential energy of this charge system.

Answer 1

Given q₁ = 5 nC = 5×10^–9 C, q₂ = – 2 nC =–2 × 10^–9 C

The charges are placed on X-axis. The distance between the charges

x = x2 –x1 = (23–5) cm = 18 cm = 0.18 m

∴ Electrostatic potential energy of charges