Question

Question: In Fig. 24-48, what is the net electric potential at the origin due to the circular arc of charge ${\mathit{Q}}_{\mathbf{1}}\mathbf{=}\mathbf{+}\mathbf{7}\mathbf{.}\mathbf{21}\mathbf{\text{pC}}$and the two particles of charges${\mathit{Q}}_{\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}{\mathit{Q}}_{\mathbf{1}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{Q}}_{\mathbf{3}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}{\mathit{Q}}_{\mathbf{1}}$? The arc’s center of curvature is at the origin and its radius R is 2. 00 m; the angle indicated is$\mathit{\theta }\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}$.

Short answer

Answer:

The net electric potential at the origin sue to the charges is 32.4 mV.

Step-by-step solution

The given data

1. Values of the charges,$Q_1=+7.21\text{pC,}{Q}_2=+4Q_{1}and{Q}_3=-2Q_1$
2. Radius of the circular arc, R = 2m
3. Angle indicated in the given figure,$\theta ={20}^0$

Understanding the concept of the electric potential

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

Calculation of the net electric potential

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

$$\begin{gathered} V_{net}=V_1+V_2+V_3 \\ =\frac{1}{4\pi {\epsilon }_0}\frac{Q_1}{R}+\frac{1}{4\pi {\epsilon }_0}\frac{Q_2}{2R}+\frac{1}{4\pi {\epsilon }_0}\frac{Q_3}{R} \\ =\frac{1}{4\pi {\epsilon }_0}\times \frac{Q_1}{R}\left[1+\frac{4}{2}-2\right] \\ =9.0\times {10}^9\times \frac{7.21\times {10}^{-12}}{2.0} \\ =32.44\times {10}^{-3}\text{V} \\ =32.4\text{mV} \end{gathered}$$

Hence, the value of the electric potential is 32.4 mV.