Two equal charges $2 \times 10^{- 7} C$ each, are held fixed at a separation of 20cm. A third charge of equal magnitude is placed midway between the charges. It is now moved to a point 20cm from both the charges. The work to be done by the electric field during the process is

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$7.2 \times 10^{- 3} J$

$72 \times 10^{- 3} J$

$3.6 \times 10^{- 3} J$

$36 \times 10^{- 3} J$

answer is A.

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Detailed Solution

$W_{e l e c t o r n} = U_{I} - U_{R}$
$= \frac{q_{1} q_{2}}{4 π ε_{0}} [\frac{2}{\frac{d}{2}} - \frac{2}{x}] = 2 \times 2 \times 10^{- 14} \times 9 \times 10^{9} (\frac{4 - 2}{0.2}) J = 36 \times 10^{- 4} J$

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