In the first configuration (1) as shown in the figure, four identical charges ($\mathrm{q}_0$) are kept at the corners $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and D of square of side length 'a'. In the second configuration (2), the same charges are shifted to mid points $\mathrm{G}, \mathrm{E}, \mathrm{H}$ and F, of the square, If $\mathrm{K}=\frac{1}{4 \pi \varepsilon_0}$, the difference between the potential energies of configuration (2) and (1) is given by :

Question

Aicharya · Accepted Answer

$\begin{aligned}
& u_{\oplus}=\left(2 \frac{K q_0}{a}+\frac{K q_0}{\sqrt{2} a}\right) q_0 \times 2 \\ & u_0=\left(2 \frac{K q_0 \sqrt{2}}{a}+\frac{K q_0}{a}\right) q_0 \times 2
\end{aligned}$
$\begin{aligned}
& \text { So, } \quad \Delta u=u_2-u_1=2 q_0 \frac{k q_0}{a}\left[2 \sqrt{2}+1-2-\frac{1}{\sqrt{2}}\right] \\ & \Rightarrow \quad \Delta u=\frac{2 q_0^2}{4 \pi \varepsilon_0 a}\left[\frac{4-\sqrt{2}-1}{\sqrt{2}}\right]=\frac{2 q_0^2}{4 \pi \varepsilon_0 a} \frac{(3-\sqrt{2})}{\sqrt{2}} \\ & \Rightarrow \quad \Delta u=\frac{2 k q_0^2}{a}\left[\frac{3-\sqrt{2}}{\sqrt{2}}\right]=\frac{k q_0^2}{a}(3 \sqrt{2}-2)
\end{aligned}$

In the first configuration (1) as shown in the figure, four identical charges (\(\mathrm{q}_0\)) are kept at…

Solution