Question

A Gaussian sphere of radius $\mathbf{4}\mathbf{.00}\mathbf{\text{cm}}$ is centered on a ball that has a radius ofrole="math" localid="1662731665361" $\mathbf{1}\mathbf{.00}\mathbf{\text{cm}}$ and a uniform charge distribution. The total (net) electric flux through the surface of the Gaussian sphere is$\mathbf{+}\mathbf{5.6}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{\text{N}}\mathbf{\cdot }{\mathbf{\text{m}}}^{\mathbf{\text{2}}}\mathbf{/}\mathbf{\text{C}}$ . What is the electric potential $\mathbf{\text{12.0cm}}$from the center of the ball?

Short answer

The electric potential from the center of the ball is $3.7\times {10}^4\text{V}$.

Step-by-step solution

Given data:

The radius of the Gaussian sphere,$\mathrm{R}=1.0\text{cm}=0.01\text{m}$

The net electric flux,$\mathrm{\varphi }=+5.6\times {10}^4\text{N}\cdot {\text{m}}^{\text{2}}/\text{C}$

The value of the electric potential at a distance,$\mathrm{R}=12.0\text{cm}=0.12\text{m}$

Understanding the concept:

Gauss's law states that the total electric flux from an enclosed surface is equal to the enclosed charge divided by the permittivity.

$\begin{array}{l}\mathbf{\varphi }\mathbf{=}\frac{\mathbf{1}}{{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{q}\\ \mathbf{}\mathbf{q}\mathbf{=}{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{\varphi }\end{array}$

And then to find potential use,

role="math" localid="1662731987630" $\begin{array}{c}\mathbf{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{o}}}\frac{\mathbf{q}}{\mathbf{R}}\mathbf{A}\\ \mathbf{=}\mathbf{k}\frac{\mathbf{q}}{\mathbf{R}}\end{array}$

Here, $\mathbf{\varphi }$ is the electric flux, ${\mathbf{\epsilon }}_{\mathbf{0}}$ is the permittivity of free space, $\mathbf{q}$ is the charge, $\mathbf{k}$ is the Coulomb’s constant having a value $\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{\text{N}}\mathbf{\cdot }{\mathbf{\text{m}}}^{\mathbf{\text{2}}}\mathbf{/}{\mathbf{\text{C}}}^{\mathbf{\text{2}}}$, and $\mathbf{\text{R}}$ is the distance,

Calculate the electric potential 12.0 cm from the center of the ball:

Write the equation for electric potential as below.

$\mathrm{V}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{\mathrm{q}}{\mathrm{R}}$

Substitute known values in the above equation.

$\begin{array}{c}\mathrm{V}=9.0\times {10}^9\times \frac{8.85\times {10}^{-12}\times 5.6\times {10}^4}{0.12}\\ =3.7\times {10}^4\text{V}\end{array}$

Hence, the electric potential from the center of the ball is $3.7\times {10}^4\text{V}$.