Figure 24-63 shows three circular, nonconducting arcs of radius$\mathbf{R}\mathbf{=}\mathbf{8}\mathbf{.50}\mathbf{cm}$ . The charges on the arcs are ${\mathbf{q}}_{\mathbf{1}}\mathbf{}\mathbf{=}\mathbf{4.52}\mathbf{\text{pC}}$,${\mathbf{q}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.00}{\mathbf{q}}_{\mathbf{1}}$ , ${\mathbf{q}}_{\mathbf{3}}\mathbf{=}\mathbf{+}\mathbf{3}\mathbf{.00}{\mathbf{q}}_{\mathbf{1}}$. With $\mathit{V}\mathbf{=}\mathbf{0}$at infinity, what is the net electric potential of the arcs at the common center of curvature?

The charge, ${\mathrm{q}}_1=4.52\text{pC}=4.52\times {10}^{-12}\text{C}$

The charge, ${\mathrm{q}}_2=-2.0{\mathrm{q}}_1$

The charge, ${\mathrm{q}}_3=+3.0{\mathrm{q}}_1$

The radius of circle$\mathrm{R}=8.50\text{cm}=0.085\text{m}$

The potential at infinity, $\mathrm{V}=0$

The potential due to a group of three charged particlesis,

$\begin{array}{c}\mathbf{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{o}}}\frac{\mathbf{1}}{\mathbf{R}}\mathbf{[}{\mathbf{q}}_{\mathbf{1}}\mathbf{+}{\mathbf{q}}_{\mathbf{2}}\mathbf{+}{\mathbf{q}}_{\mathbf{3}}\mathbf{]}\\ \mathbf{=}\frac{\mathbf{k}}{\mathbf{R}}\mathbf{[}{\mathbf{q}}_{\mathbf{1}}\mathbf{+}{\mathbf{q}}_{\mathbf{2}}\mathbf{+}{\mathbf{q}}_{\mathbf{3}}\mathbf{]}\end{array}$

Here, $\mathbf{V}$ is the electric potential, ${\mathbf{\epsilon }}_{\mathbf{0}}$ is the permittivity of free space, $\mathbf{R}$ is the distance, ${\mathbf{q}}_{\mathbf{1}}$, ${\mathbf{q}}_{\mathbf{2}}$, and ${\mathbf{q}}_{\mathbf{3}}$ are the charges, $\mathbf{k}$is the Coulomb’s constant having a value of $\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}{\mathbf{\text{Nm}}}^{\mathbf{\text{2}}}\mathbf{/}{\mathbf{\text{C}}}^{\mathbf{\text{2}}}$

Draw the figure as below.

The net electric potential at the centre of the curvature is given as:

$\begin{array}{c}\mathrm{V}={\mathrm{V}}_1+{\mathrm{V}}_2+{\mathrm{V}}_3\\ =\frac{\mathrm{k}}{\mathrm{R}}[{\mathrm{q}}_1+{\mathrm{q}}_2+{\mathrm{q}}_3]\\ =\frac{\mathrm{k}}{\mathrm{R}}[{\mathrm{q}}_1+(-2.0{\mathrm{q}}_1)+(+3.0{\mathrm{q}}_1)]\end{array}$

$\begin{array}{c}\mathrm{V}=\frac{9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2}{0.085\text{m}}[(4.52\times {10}^{-12}\text{C})-(2.0\times 4.52\times {10}^{-12}\text{C})+(3.0\times 4.52\times {10}^{-12}\text{C})]\\ =\text{105.8}\times {10}^9\text{N}\cdot \text{m}/{\text{C}}^2[(4.52\times {10}^{-12}\text{C})-(\text{9.04}\times {10}^{-12}\text{C})+(\text{13.56}\times {10}^{-12}\text{C})]\\ =\text{105.8}\times {10}^9\left[(\text{9.04}\times {10}^{-12})\right]\text{V}\\ =0.956\text{V}\end{array}$

Hence, the net electric potential of the arcs at the common center of curvature is $0.956\text{V}$.