Charges pf $10 μ C, 20 μ C, a n d - 30 μ C$ are kept at the vertices of an equilateral triangle of side 3 m. What energy will be required to dissociate the assembly of charges?

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Solution

Total potential energy of the system-

$U = U_{A B} + U_{B C} + U_{A C}$

$⟹ U = \frac{k q_{A} q_{B}}{a} + \frac{k q_{B} q_{C}}{a} + \frac{k q_{A} q_{C}}{a}$

$⟹ U = \frac{k}{a} (q_{A} q_{B} + q_{B} q_{C} + q_{A} q_{c})$

$⟹ U = \frac{9 \times 10^{9}}{3} (200 - 600 - 300) \times 10^{- 12}$

$⟹ U = 3 \times 10^{9} \times (- 700) \times 10^{- 12} J$

$⟹ U = - 2.1 J$

When they are dissociated, there will be no potential energy stored-

$⟹ U^{'} = 0$

Hence , work done $W = U^{'} - U$

$⟹ W = 2.1 J$

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