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.