91Ó°ÊÓ

1. Uniformly distributed charge on the rod,$Q=-25.6\text{pC}$
2. Radius of the circular arc,R = 3.71 cm
3. Central angle of the circular arc,$\mathrm{Ï•}={120}^0$

The electric potential at infinity is V = 0

Integrating the basic formula of the potential expression of the rod, we can get the value of the electric potential using the substituted values.

Formula:

The expression for the electric potential of a point particle, $V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q}{r}$ (i)

Here, q is the charge, r is the distance between the point and the charge, and is the permittivity of free space.

Consider a small portion in the rod with small charge, from the given figure of the rod, we can get the electric potential at P using equation (i) as follows:

(In the figure, is the radius of the arc, is the point where the electric potential to be calculated, is the charge in the small portion, and is the angle of the given arc.)

.$$\begin{gathered} {V=∫\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{dq}{R} \\ =\frac{1}{\left(4\mathrm{Ï€}{\mathrm{Î\mu }}_0\right)\left(R\right)}∫dq \\ =\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q}{R} \\ =\left(9×{10}^9{\text{N.m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(\left(-25.6\text{pC}\right)\left(\frac{{10}^{-12}\text{C}}{1\text{pC}}\right)\right)}{\left(3.71\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)} \\ =-6.21\text{V} \\ }_{} \end{gathered}$$

Therefore, the electric potential at point P due to charged rod is -6.21 V.