Question

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?

Short answer

The net electric potential of the arcs at the common center of curvature is $\mathrm{V}=0.956\text{V}$.

Step-by-step solution

 Step 1: Given data:

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, $\mathrm{V}=0$

Understanding the concept:

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.

Calculate the net electric potential of the arcs at the common center of curvature:

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 $0.956\text{V}$.