Question

(a) Deduce the expression for the potential energy of a system of two charges $q_1$ and $q_2$ located $\overrightarrow{r_1}$ and $\overrightarrow{r_2}$, respectively, in an external electric field.
(b) Three point charges, + Q + 2Q and - 3Q are placed at the vertices of an equilateral triangle ABC of side l. If these charges are displaced to the mid-point $A_1,B_1$ and $C_1$, respectively, find the amount of the work done in shifting the charges to the new locations.

Answer 1

$\left(a\right)\to \left(i\right)$ work done to bring a charge q in the electric field at a distance $r_1=q_{1}V\left(r_1\right)$

$\left(ii\right)$ work done to bring a charge q in the electric field at a distance $r_2=q_{2}V\left(r_2\right)$

$\left(iii\right)$ work done on $q_2$ to move it against the force of $q_1=\frac{K{q}_1{q}_2}{r_{12}}$

$\therefore$ The potential energy of the system $=q_{1}V\left(r_1\right)+q_2\left(V\right)\left(r_2\right)+\frac{K{q}_1{q}_2}{r_{12}}$

$\left(b\right)$ work done in moving charges $=$ Potential of the initial position - Potential of the final position.

Now, Potential of the initial system $=V_{12}+V_{23}+V_{31}={2KQ}^2+\left({-3KQ}^2\right)+(-6K{Q}^2)=-7K{Q}^{2}J$

potential energy after moving to position, $A_1,B_1\&C_1$

$=\frac{K\left(2Q\right)\left(Q\right)}{1/2}+\frac{K\left(Q\right)(-3Q)}{1/2}+\frac{K(-3Q)\left(2Q\right)}{1/2}$

$=4{KQ}^2-6{KQ}^2-12{KQ}^2=-14K{Q}^{2}J$

$\therefore$ work done $=-7Q^{2}K-(-14K{Q}^2)=7K{Q}^{2}J$