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Two charged particles, having equal charges of $2.0 \times 10^{–5} C$ each, are brought from infinity to within a separation of $10\; cm$. Find the increase in the electric potential energy during the process.

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Here, $q_1 = q_2 = 2.0 \times 10^{–5} C $
$r= 10 cm, 10^{-1}m$
Electric potential energy at infinite separation is zero.
Therefore, increase in P.E. = potential energy at this distance
$\qquad= \large\frac{q_1q_e}{4 \pi \in_0 r} =\frac{9 \times 10^9 (2 \times 10^{-5} )(2 \times 10^{-5})}{10^{-1}}$
$\qquad= 36 joule$
Hence B is the correct answer.
answered Jun 11, 2014 by meena.p

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