91Ó°ÊÓ

1. The radius of the full circle is$R_1=6\text{cm}$
2. The radius of the second thin rod is$R_2=4\text{cm}$
3. Charge of the thin rod that forms full circle of radius,$R_1=6\text{cm}$ is $q_1=-3\mathrm{μ}\text{C}$.
4. Charge of the second thin rod of radius$R_2=4\text{cm}$, is $q_2=2\mathrm{μ}\text{C}$that subtends an angle of 90⁰ about the center of the circle.

Magnitude of the dipole moment that is perpendicular to a radial line,$p=1.28×{10}^{-21}\text{C}·\text{m}$

Using the concept of the electric potential, we can get the net electric potential at the point due to charges can be calculated by adding all the potentials due to the individual charges.

Formula:

The electric potential at point P due to the circle is given by, $V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q}{R}$ (i)

The electric potential due to first charge is given using equation (i) and the given values as follows:

$$\begin{gathered} V_1=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q_1}{R_1} \\ =9.0×{10}^9×\frac{\left(-3\mathrm{μ}\text{C}\right)}{\left(0.06\text{m}\right)} \\ =-4.5×{10}^5\text{V} \end{gathered}$$

Due to the arc, the potential of the second charge at P is given using the given data in equation (i) as follows:

$$\begin{gathered} V_2=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q_2}{R_2} \\ =9.0×{10}^9×\frac{2×{10}^{-6}\text{C}}{0.04\text{m}} \\ =+4.5×{10}^5\text{V} \end{gathered}$$

The potential at P due to the dipole at P is V₃= 0 (at equatorial position)

Thus, the net electric potential at P using the above values is given as:

$$\begin{gathered} V_{net}=V_1+V_2+V_3 \\ =-4.5×{10}^5+4.5×{10}^5+0 \\ =0\text{V} \end{gathered}$$

Hence, the net electric potential is 0V.