First slide

Potential energy of system of charge in external field

Question

Two particles each having charge of $5 μC$ are placed at a separation of 1m. what would be the work done by external force in pulling them apart slowly by 2m, so that new separation becomes 3m?

Moderate

Solution

$\begin{array}{l} From work energy theorem, \\ ΔK = W_{ext} + W_{int} = W_{ext} - W_{ele} \\ 0 = W_{ext} - (U_{f} - U_{i}) \\ W_{ext} = \frac{q_{1} q_{2}}{4 {πε}_{0}} (\frac{1}{r_{2}} - \frac{1}{r_{1}}) = {(5 \times 10^{- 6})}^{2} \times 9 \times 10^{9} (\frac{1}{3} - 1) = - 0 .15 J \end{array}$

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