A charge of$\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}\mathbf{C}$lies on an isolated metal sphere of radius 16.0 cm. With $\mathbf{V}\mathbf{=}\mathbf{0}$at infinity, what is the electric potential at points on the sphere’s surface?

For the isolated metal sphere:

The radius, $\mathrm{R}=16.0\mathrm{cm}$

The value of the charge $\mathrm{q}=1.50\times {10}^{-8}\mathrm{C}$

At infinity, $\mathrm{V}=0$

Use the equation of the electric potential at points on the sphere’s surface.

The electric potential is defined as the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field.

Formula:

The electric potential is define by,

$\begin{array}{rcl}\mathrm{V}& =& \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{\mathrm{R}}\\ & =& \mathrm{k}\frac{\mathrm{q}}{\mathrm{R}}\end{array}$

Here, V is electric potential, R is distance between the point charges, q is charge, ${\mathrm{\epsilon }}_0$is the permittivity of free space, and k is the Coulomb’s constant having a value $9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2$.

The electric potential on the sphere is,

$\begin{array}{rcl}\mathrm{V}& =& \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{\mathrm{R}}\\ & =& \mathrm{k}\frac{\mathrm{q}}{\mathrm{R}}\end{array}$

Substitute known values in the above equation.

$\begin{array}{rcl}\mathrm{V}& =& 9.0\times {10}^9\times \frac{1.5\times {10}^{-8}}{0.16}\\ & =& 843.7\mathrm{volts}\\ & \approx & 844\mathrm{volts}\end{array}$

Hence, the electric potential at points on the sphere’s surface is 844 V.