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Question

Four charges +q, -q, +q and -q are placed in order on the four consecutive corners of a square of side a. Find the work done in interchanging the positions of any two neighbouring charges of opposite sign.

Solution
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Let $$ U_1 $$ be the potential energy of system I and $$ U_{11} $$ be the potential energy of system II.
Work done = change in P.E.= $$ U_{11} -U_1 $$
$$ U_1 = - \frac {4q^2}{ 4\pi \epsilon_0 a} + \frac {2q^2}{ 4 \pi \epsilon_0 \sqrt {2}a } $$
$$ U_{11} = \frac { -2q^2}{ 4 \pi \epsilon_0 \sqrt {2} a} $$
$$ \therefore W = \frac { -2q^2}{ 4 \pi \epsilon_0 \sqrt {2} a} + \frac { 4q^2}{ 4 \pi \epsilon_0 a } - \frac { 2q^2}{ 4 \pi \epsilon_0 \sqrt {2} a } $$
$$ = \frac {q^2}{ 4 \pi \epsilon_0 a} ( 4 - 2 \sqrt {2} ) $$

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