c
From figure, \(A C=L, B C=L, B D=B C=L\)

\(A D=A B+B D=2 L+L=3 L\)

Potential at \(C\) is given by

\(V_{C}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{A C}+\frac{(-q)}{B C}\right]=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{L}-\frac{q}{L}\right]=0\)

Potential at \(D\) is given by

\(V_{D} =\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{A D}+\frac{(-q)}{B D}\right]=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{3 L}-\frac{q}{L}\right]\)

\(=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{L}\left[\frac{1}{3}-1\right]=\frac{-q}{6 \pi \varepsilon_{0}}\)

Work done in moving charge \(+Q\) along the semicircle \(CRD\) is given by

\(W=\left[V_{D}-V_{C}\right](+Q)=\left[\frac{-q}{6 \pi \varepsilon_{0}}-0\right](Q)=\frac{-q Q}{6 \pi \varepsilon_{0} L}\)

Comments : Potential at \(C\) is zero because the charges are equal and opposite and the distances are the same. Potential at \(D\) due to \(-q\) is greater than that at \(A\) \((+q),\) because \(D\) is closer to \(B .\) Therefore it is negative.