Answer
$V = \dfrac{1}{4 \pi \epsilon_{0}} \dfrac{Q}{a}$
Work Step by Step
The potential across the arc for a segment is given by
\begin{equation}
d V=\frac{1}{4 \pi \epsilon_{0}} \frac{d q}{r}
\end{equation}
Where $dq$ is very small to be a line. So, it is equal to
$$\lambda d l = \dfrac{Q}{0.5 (2 \pi a)} = \dfrac{Q}{ \pi a}$$.
The term $0.5 (2 \pi a)$ is the circumference of the semicircle.
Also, $dl = a d\theta$.
Now we integrate over this segment to get the potential over the arc to get the potential
$$V=\frac{1}{4 \pi \epsilon_{0}} \int_{0}^{\pi} \frac{Q d \theta}{\pi a}=\boxed{\frac{1}{4 \pi \epsilon_{0}} \frac{Q}{a}}$$
