91Ó°ÊÓ

Given data:

The radius of the semi-circular arc is (a) and the net electric charge on the rod is Q.

Let taking a segment of the rod having the length as (dl)and charge(dq).

So, the potential is,

$V=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}∫\frac{dq}{r}$

Here r is the distance therefore r = a

As the chargedqin the segment dlwill be given by linear charge density $\mathrm{λ}$

$dq=\mathrm{λ}dl$

Taking infinitesimal triangle, the segment dlto the circle having radius (a)

$dl=ad\mathrm{θ}$

$\mathrm{λ}=\frac{Q}{\mathrm{Ï€²¹}}$By using the charge (Q) distributed on the rod (semicircle) linear charge density$\left(\mathrm{λ}\right)$

Hence, the expression for $\mathrm{λ}$and dlis

$$\begin{gathered} dq=\mathrm{λ}dl \\ dq=\left(\frac{Q}{\mathrm{π}a}\right)\left(ad\mathrm{θ}\right) \\ dq=\frac{Qd\mathrm{θ}}{\mathrm{π}} \end{gathered}$$

Now the potential at the center of the semi-circle from is

$$\begin{gathered} V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{°}}∫\frac{dq}{a} \\ =\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{°}}{∫}_0^{\mathrm{Ï€}}\frac{Qd\mathrm{θ}}{a} \\ =\frac{Q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{°}}\frac{[\mathrm{Ï€}-0]}{\mathrm{Ï€}a} \\ =\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{°}}\frac{Q}{a} \end{gathered}$$

Hence, the potential at the center of curvature is localid="1664272157987" $V=\frac{1}{4{\mathrm{π}}_{°}}\frac{Q}{a}$.