Determine the electrostatic potential energy of a system consisting of two charges $7 μ C$ and $- 2 μ C$ separated by distance of 20 cm.

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Solution

Electric potential energy, $U = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{12}}$

Given, $q_{1} = 7 \times 10^{- 6} C$

$q_{2} = - 2 \times 10^{- 6} C$

$r_{12} = 20 c m = 0.2 m$

So,

$U = \frac{9 \times 10^{9} \times 7 \times 10^{- 6} \times 2 \times 10^{- 6}}{0.2}$

$U = - 630 \times 10^{- 3} J$

$U = - 0.63 J$

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Determine the electrostatic potential energy of a system consisting of two charges 7 μC and −2 μC separated by distance of 20 cm.

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