Question

Question: In Fig. 24-46, three thin plastic rods form quarter-circles with a common center of curvature at the origin. The uniform charges on the three rods are,${\mathit{Q}}_{\mathbf{1}}\mathbf{=}\mathbf{+}\mathbf{30}\mathbf{\text{nC,}}{\mathit{Q}}_{\mathbf{2}}\mathbf{=}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}{\mathit{Q}}_{\mathbf{1}}\mathbf{\text{and}}{\mathit{Q}}_{\mathbf{3}}\mathbf{=}\mathbf{-}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}{\mathit{Q}}_{\mathbf{1}}$. What is the net electric potential at the origin due to the rods?

Short answer

Answer:

The net electric potential at the origin due to the rods is +13.5 KV.

Step-by-step solution

The given data

1. The uniform charges on the three rods,$Q_1=+30\text{nC,}{Q}_2=+90\text{nCand}{Q}_1=-240\text{nC}$.
2. The different radii are given as:$R_3=0.04\text{m,}{R}_2=0.02\text{m,}{R}_1=0.01\text{m}$

Understanding the concept of the electric potential

Using the concept of electric potential at a point due to many charges present in the system, we can get the required electric potential at the given point at a distance from the center of the y-axis.

Formula:

The electric potential at a point due to a charge,$V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}$ (i)

Calculation of the net electric potential

The net electric potential at the origin due to the charges present in the system is given using equation (i) as follows:

$$\begin{gathered} V_{net}=V_1+V_2+V_3 \\ {V}_{net}=\frac{1}{4\pi {\epsilon }_0}\left[\frac{Q_3}{R_3}+\frac{Q_2}{R_2}+\frac{Q_1}{R_1}\right] \\ =9.0\times {10}^9\times Q_1\left[\frac{-8}{0.04\text{m}}+\frac{3}{0.02\text{m}}+\frac{1}{0.01\text{m}}\right] \\ =9.0\times {10}^9\times 30\times {10}^{-9}\left[-200+150+100\right] \\ =+1.35\times {10}^4\text{V} \\ =+13.5\text{kV} \end{gathered}$$

Hence, the net potential is +13 .5 KV.