Question

Derive the expression for the potential energy of a system of two charges in the absence of the external electric field.

Answer 1

Hint: The potential energy of this system of charge is equal to total work done ,i.e., To move charge q${}_1$ from infinity to A and charge q${}_2$ from infinity to B. when we bring charge q${}_2$ from infinity to point B, q${}_1$ is also taken into account, whereas in case of q${}_1$, the charge q${}_2$ is not taken because there is no initial electric field.

Complete step-by-step answer:

The work done to move q from infinity to A is, $W_1$
($\therefore$ There is no initial electric field hence work done to bring charge q${}_1$ from infinity to A is zero)
The work done to move q${}_2$ from infinity to B is, $$W_1=V_1{q}_1$$
Here,$V_1$ is the electric potential at B due to $q_1$ , it is given by:
$V_1={\displaystyle \frac{1}{4\pi {\xi }_{\circ }}}\left({\displaystyle \frac{q_1}{r_{12}}}\right)$
$\Rightarrow W_2=V_1{q}_2=V_1={\displaystyle \frac{1}{4\pi {\xi }_{\circ }}}\left({\displaystyle \frac{q_1}{r_{12}}}\right)q_2$
$W_2={\displaystyle \frac{1}{4\pi {\xi }_{\circ }}}\left({\displaystyle \frac{q_1{q}_2}{r_{12}}}\right)$
The potential energy of this system of charge is equal to total work done to bring the charges from infinity to A or B.
$$U=W_1+W_2=0+{\displaystyle \frac{1}{4\pi {\xi }_{\circ }}}\left({\displaystyle \frac{q_1{q}_2}{r_{12}}}\right)$$
$U=\left({\displaystyle \frac{1}{4\pi {\xi }_{\circ }}}\right)\left({\displaystyle \frac{q_1{q}_2}{r{}_{12}}}\right)$
$U=K\left({\displaystyle \frac{q_1{q}_2}{r_{12}}}\right)$. (Where,$K=\left({\displaystyle \frac{1}{4\pi {\xi }_{\circ }}}\right)$ )

Note: Generally students can go wrong in considering the work done to move charge q${}_1$ from infinity to A where there is no external electric field working, the same goes for charge q${}_2$ where actually the charge q${}_1$ affects the work done. We can also consider q${}_2$ as the first charge which comes into the system from infinity in that case q${}_1$ will not affect the work done to bring the charge from infinity to B, but q${}_2$ will affect work done to bring charge $q_1$ to A.