Answer:
[tex]v = \frac{kQ}{a}[/tex]
Explanation:
We define the linear density of charge as:
[tex]\lambda = \frac{Q}{L}[/tex]
Where L is the rod's length, in this case the semicircle's length L = πr
The potential created at the center by an differential element of charge is:
[tex]dv = \frac{kdq}{r}[/tex]
where k is the coulomb's constant
r is the distance from dq to center of the circle
Thus.
[tex]v = \int_{}^{}\frac{kdq}{a}[/tex]
[tex]v = \frac{k}{a}\int_{}^{}dq[/tex]
[tex]v = \frac{kQ}{a}[/tex] Potential at the center of the semicircle
