Four equal charges Q are placed at the four c | vedclass

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Question

(c) Potential at centre $O$ of the square

${V_O} = 4\,\left( {\frac{Q}{{4\pi {\varepsilon _0}(a/\sqrt 2 )}}} \right)$

Work done in shifting $(-Q

Vedclass · Accepted Answer

$\frac{{\sqrt 2 {Q^2}}}{{\pi {\varepsilon _0}a}}$

Vedclass · Answer

(c) Potential at centre $O$ of the square

${V_O} = 4\,\left( {\frac{Q}{{4\pi {\varepsilon _0}(a/\sqrt 2 )}}} \right)$

Work done in shifting $(-Q)$ charge from centre to infinity $W = - \,Q({V_\infty } - {V_O}) = Q{V_0}$$ = \frac{{4\sqrt 2 \,{Q^2}}}{{4\pi {\varepsilon _0}a}}$$ = \frac{{\sqrt 2 {Q^2}}}{{\pi {\varepsilon _0}a}}$

Four equal charges $Q$ are placed at the four corners of a square of each side is $'a'$. Work done in removing a charge $-Q$ from its centre to infinity is

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