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2. Electric Potential and Capacitance
medium
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
A
$0$
B
$\frac{{\sqrt 2 {Q^2}}}{{4\pi {\varepsilon _0}a}}$
C
$\frac{{\sqrt 2 {Q^2}}}{{\pi {\varepsilon _0}a}}$
D
$\frac{{{Q^2}}}{{2\pi {\varepsilon _0}a}}$
(AIIMS-1995)
Solution
(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}}$
Standard 12
Physics
