Question

Obtain an expression for the work done to dissociate the system of three charges placed at the vertices of an equilateral triangle of side ‘a’ as shown below.

 

Answer 1

Work done to keep the system bound is

${\displaystyle W=\frac{K{q}_1{q}_2}{a}+\frac{K{q}_2{q}_3}{a}+\frac{K{q}_1{q}_3}{a}}$

${\displaystyle W=\frac{Kq(-4q)}{a}+\frac{K(-4q)2q}{a}+\frac{Kq\left(2q\right)}{a}}$

${\displaystyle W=\frac{-K4q^2}{a}-\frac{K8q^2}{a}+\frac{K2q^2}{a}}$

${\displaystyle W=\frac{K{q}^2}{a}(-4-8+2)=\frac{-10K{q}^2}{a}}$

Therefore, the work done to dissociate the system is

${\displaystyle W_d=-W=\frac{10K{q}^2}{a}}$