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(C) The potential energy $U$ of a system of three charges $q$ at the vertices of an equilateral triangle of side $r$ is given by $U = 3 	imes \frac{1

Vedclass · Accepted Answer

$\frac{1}{4 \pi \varepsilon_0} \frac{3 q^2}{L}$

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(C) The potential energy $U$ of a system of three charges $q$ at the vertices of an equilateral triangle of side $r$ is given by $U = 3 	imes \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r}$.Initial potential energy $U_i$ with side $L$ is $U_i = \frac{3}{4 \pi \varepsilon_0} \frac{q^2}{L}$.Final potential energy $U_f$ with side $\frac{L}{2}$ is $U_f = \frac{3}{4 \pi \varepsilon_0} \frac{q^2}{L/2} = \frac{6}{4 \pi \varepsilon_0} \frac{q^2}{L}$.The work done $W$ is equal to the change in potential energy: $W = U_f - U_i$.$W = \frac{6}{4 \pi \varepsilon_0} \frac{q^2}{L} - \frac{3}{4 \pi \varepsilon_0} \frac{q^2}{L} = \frac{3}{4 \pi \varepsilon_0} \frac{q^2}{L}$.

Three particles of each charge $q$ are placed at the vertices of an equilateral triangle of side $L$. The work to be done to decrease the side of the triangle to $\frac{L}{2}$ is

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