First slide

Potential energy

Question

The work done required to put the four charges together at the corners of a square as shown in fig. is

Easy

A

$\frac{1}{4 {πε}_{0}} \times \frac{q^{2}}{a}$
B

$- \frac{2 \cdot 6}{4 {πε}_{0}} \times \frac{q^{2}}{a}$
C

$\frac{2 \cdot 6}{4 {πε}_{0}} \times \frac{q^{2}}{a}$
D

none of these

Solution

See fig. The work done required is equal to the potential energy of the system. Thus
$W = U = \frac{1}{4 {πε}_{0}} [\frac{q_{1} q_{2}}{r_{12}} + \frac{q_{2} q_{3}}{r_{23}} + \frac{q_{3} q_{4}}{r_{34}} + \frac{q_{4} q_{1}}{r_{41}} + \frac{q_{1} q_{3}}{r_{13}} + \frac{q_{2} q_{4}}{r_{24}}]$
$= \frac{1}{4 {πε}_{0}} [- \frac{q^{2}}{a} - \frac{q^{2}}{a} - \frac{q^{2}}{2} - \frac{q^{2}}{2} + \frac{q^{2}}{a \sqrt{2}} + \frac{q^{2}}{a \sqrt{2}}] \begin{array}{l} = \frac{1}{4 {πε}_{0}} [- \frac{4 q^{2}}{a} + \frac{2 q^{2}}{a \sqrt{2}}] \\ = \frac{q^{2}}{- 4 {πε}_{0} a} [4 - \sqrt{2}] = - \frac{2 \cdot 6}{4 {πε}_{0}} \times \frac{q^{2}}{a} \end{array}$

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