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The work done required to put the four charges together at the corners of a square of side $a$ , as shown in the figure is
$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q^2}}}{a}$
$\frac{{ - 2.6}}{{4\pi {\varepsilon _0}}}\frac{{{q^2}}}{a}$
$ + \frac{{2.6}}{{4\pi {\varepsilon _0}}}\frac{{{q^2}}}{a}$
none of these
Solution
The work done required is equal to the potential energy of the system. Thus,
$ W= \mathrm{U}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{12}}+\frac{\mathrm{q}_{2} \mathrm{q}_{3}}{\mathrm{r}_{23}}+\frac{\mathrm{q}_{3} \mathrm{q}_{4}}{\mathrm{r}_{34}}+\frac{\mathrm{q}_{4} \mathrm{q}_{1}}{\mathrm{r}_{41}}+\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{\mathrm{r}_{13}}+\frac{\mathrm{q}_{2} \mathrm{q}_{2}}{\mathrm{r}_{24}}\right]$
$=\frac{1}{4 \pi \varepsilon_{0}}\left[-\frac{4 \mathrm{q}^{2}}{\mathrm{a}}+\frac{2 \mathrm{q}^{2}}{\mathrm{a} \sqrt{2}}\right]$
$=-\frac{\mathrm{q}^{2}}{4 \pi \varepsilon_{0} \mathrm{a}}[4-\sqrt{2}]=-\frac{2.6}{4 \pi \varepsilon_{0}}\left(\frac{\mathrm{q}^{2}}{\mathrm{a}}\right) $
