Charges +q and -q are placed at points A and | vedclass

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(C) From the figure,$AC = L$,$BC = L$. Since $C$ is the midpoint of $AB$ and $CD$ is a semicircle with diameter $CD$,the distance $BD = L$ (as $C, B,

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

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(C) From the figure,$AC = L$,$BC = L$. Since $C$ is the midpoint of $AB$ and $CD$ is a semicircle with diameter $CD$,the distance $BD = L$ (as $C, B, D$ are collinear and $CD$ is the diameter of the semicircle,$CB = BD = L$).Potential at $C$ is given by:$V_C = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{AC} + \frac{-q}{BC} \right] = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{L} - \frac{q}{L} \right] = 0$Potential at $D$ is given by:$AD = AB + BD = 2L + L = 3L$$V_D = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{AD} + \frac{-q}{BD} \right] = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{3L} - \frac{q}{L} \right]$$V_D = \frac{q}{4\pi\varepsilon_0 L} \left[ \frac{1}{3} - 1 \right] = \frac{q}{4\pi\varepsilon_0 L} \left( -\frac{2}{3} \right) = -\frac{q}{6\pi\varepsilon_0 L}$Work done in moving charge $+Q$ from $C$ to $D$ is:$W = Q(V_D - V_C) = Q \left( -\frac{q}{6\pi\varepsilon_0 L} - 0 \right) = -\frac{qQ}{6\pi\varepsilon_0 L}$

Charges $+q$ and $-q$ are placed at points $A$ and $B$ respectively,which are a distance $2L$ apart. $C$ is the midpoint between $A$ and $B$. The work done in moving a charge $+Q$ along the semicircle $CRD$ is

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