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(C) The given situation is shown in the figure. The distance $AC = CB = L$. The point $D$ is at a distance $L$ from $B$,so $AD = 3L$ and $BD = L$.Work

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$-\frac{qQ}{6\pi\varepsilon_0 L}$

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(C) The given situation is shown in the figure. The distance $AC = CB = L$. The point $D$ is at a distance $L$ from $B$,so $AD = 3L$ and $BD = L$.Work done $W$ in moving a charge $Q$ along the semi-circle $CRD$ is equal to the change in potential energy of the system.$W = U_{	ext{final}} - U_{	ext{initial}}$$U_{	ext{initial}}$ is the potential energy when $Q$ is at $C$:$U_{	ext{initial}} = \frac{kqQ}{AC} + \frac{k(-q)Q}{BC} + \frac{kq(-q)}{AB} = \frac{kqQ}{L} - \frac{kqQ}{L} - \frac{kq^2}{2L} = -\frac{kq^2}{2L}$$U_{	ext{final}}$ is the potential energy when $Q$ is at $D$:$U_{	ext{final}} = \frac{kqQ}{AD} + \frac{k(-q)Q}{BD} + \frac{kq(-q)}{AB} = \frac{kqQ}{3L} - \frac{kqQ}{L} - \frac{kq^2}{2L} = -\frac{2kqQ}{3L} - \frac{kq^2}{2L}$$W = (-\frac{2kqQ}{3L} - \frac{kq^2}{2L}) - (-\frac{kq^2}{2L}) = -\frac{2kqQ}{3L}$Substituting $k = \frac{1}{4\pi\varepsilon_0}$:$W = -\frac{2}{3} \cdot \frac{1}{4\pi\varepsilon_0} \cdot \frac{qQ}{L} = -\frac{qQ}{6\pi\varepsilon_0 L}$

Electric charges $+q$ and $-q$ are placed at points $A$ and $B$ respectively,which are at a distance of $2L$ apart. If $C$ is the midpoint between $A$ and $B$,then the work done in moving a charge $+Q$ along the semi-circle $CRD$ is

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