In the first configuration (1) as shown in the figure, four identical charges ( $q_{0}$ ) are kept at the corners A, B, C and D of square of side length 'a'. In the second configuration (2), the same charges are shifted to mid points G, E, H and F, of the square. If $K = \frac{1}{4 π ε_{0}}$ , the difference between the potential energies of configuration (2) and (1) is given by [2025]

Question

Smart Achievers · Accepted Answer

(4)

$q_{0}$

$U_{2} = (2 \frac{K q_{0} \sqrt{2}}{a} + \frac{K q_{0}}{a}) q_{0} \times 2$

So, $△ U = U_{2} - U_{1} = 2 q_{0} \frac{K q_{0}}{a} [2 \sqrt{2} + 1 - 2 - \frac{1}{\sqrt{2}}]$

$\Rightarrow △ U = \frac{2 q_{0}^{2}}{4 π ε_{0} a} [\frac{4 - \sqrt{2} - 1}{\sqrt{2}}] = \frac{2 q_{0}^{2}}{4 π ε_{0} a} \frac{(3 - \sqrt{2})}{\sqrt{2}}$

$\Rightarrow △ U = \frac{2 K q_{0}^{2}}{a} [\frac{3 - \sqrt{2}}{\sqrt{2}}] = \frac{K q_{0}^{2}}{a} (3 \sqrt{2} - 2)$

Solution

1 $\frac{K q_{0}^{2}}{a} (4 \sqrt{2} - 2)$

2 $\frac{K q_{0}^{2}}{a} (3 - \sqrt{2})$

3 $\frac{K q_{0}^{2}}{a} (4 - 2 \sqrt{2})$

4 $\frac{K q_{0}^{2}}{a} (3 \sqrt{2} - 2)$

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Solution

1 Kq02a(42–2)

2 Kq02a(3–2)

3 Kq02a(4–22)

4 Kq02a(32–2)

Ans. (4) U1=(2Kq0a+Kq02a)q0×2 U2=(2Kq02a+Kq0a)q0×2 So, △U=U2–U1=2q0Kq0a[22+1–2–12] ⇒ △U=2q024πε0a[4–2–12]=2q024πε0a(3–2)2 ⇒ △U=2Kq02a[3–22]=Kq02a(32–2)

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1 $\frac{K q_{0}^{2}}{a} (4 \sqrt{2} - 2)$

2 $\frac{K q_{0}^{2}}{a} (3 - \sqrt{2})$

3 $\frac{K q_{0}^{2}}{a} (4 - 2 \sqrt{2})$

4 $\frac{K q_{0}^{2}}{a} (3 \sqrt{2} - 2)$