Question

What is the excess charge on a conducting sphere of radius $\mathbf{r}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.15}\mathbf{\text{m}}$if the potential of the sphere is$\mathbf{1500}\mathbf{\text{V}}$ and at infinity$\mathbf{V}\mathbf{=}\mathbf{}\mathbf{0}$ ?

Short answer

The excess charge on a conducting sphere is $\mathrm{Q}=2.5\times {10}^{-8}\text{C}$.

Step-by-step solution

Given data:

The potential of the sphere, $\mathrm{V}=1500\text{V}$

The permittivity of free space, ${\mathrm{\epsilon }}_0=8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}$

The radius of the conducting sphere, $\mathrm{r}=0.15\text{m}$

Understanding the concept:

Use the formula for the potential inside a sphere. The electric potential due to a sphere of radius $\mathbf{r}$is,

$\mathbf{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{Q}}{\mathbf{r}}$

Here${\mathbf{\epsilon }}_{\mathbf{0}}$is permittivity of free space,$\mathbf{Q}$is the charge, and $\mathbf{r}$is the radius of the sphere.

Rearrange the equation for the charge on the sphere.

$\mathbf{Q}\mathbf{=}\mathbf{V}\mathbf{\left(}\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{\right)}\mathbf{r}$

Calculate the excess charge on a conducting sphere:

Write the equation for the charge as below.

$\mathrm{Q}=\mathrm{V}\left(4{\mathrm{\pi \epsilon }}_0\right)\mathrm{r}$

Substitute known values in the above equation.

$\begin{array}{c}\mathrm{Q}=\left(1500\text{V}\right)(4\times 3.14(8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}))\left(0.15\text{m}\right)\\ =2.5\times {10}^{-8}\text{C}\end{array}$

Hence, the excess charge on the sphere is $2.5\times {10}^{-8}\text{C}$ .