Question

Figure 24-68 shows a hemisphere with a charge of $\mathbf{\text{4.00μC}}$ distributed uniformly through its volume. The hemisphere lies on an xy plane the way half a grapefruit might lie face down on a kitchen table. Point P is located on the plane, along a radial line from the hemisphere’s center of curvature, at radial distance$\mathbf{\text{15cm}}$ .What is the electric potential at point P due to the hemisphere?

Short answer

The electric potential at point P is$\mathrm{V}=\text{0.24MV}$

Step-by-step solution

Step 1: Given data:

The charge on the hemisphere,$\mathrm{q}=4.0\text{μC}=4.0\times {10}^{-6}\text{C}$.

The radius of the curvature, $\mathrm{R}=15\text{cm}=\text{0.15m}$.

Determining the concept:

The electric potential at point P will be the same as on its surface because the charges are uniformly distributed.

Formula:

The potential is define by,

$\begin{array}{c}\mathrm{V}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\cdot \frac{\mathrm{q}}{\mathrm{R}}\\ =\mathrm{k}\frac{\mathrm{q}}{\mathrm{R}}\end{array}$

Here, ${\epsilon }_0$ is the permittivity of free space, $\mathrm{q}$ is the electron charge, and $\mathrm{R}$ is the distance, and $\mathrm{k}$ is the Coulomb’s constant having a value $9\times {10}^9{\text{Nm}}^2/{\text{C}}^2$.

Determining the electric potential:

Write the equation for the potential as below.

$\mathrm{V}=\mathrm{k}\cdot \frac{\mathrm{q}}{\mathrm{R}}$

Substitute known values in the above equation.

$\begin{array}{c}\mathrm{V}=(9\times {10}^9{\text{Nm}}^2/{\text{C}}^2)\times \frac{4.0\times {10}^{-6}\text{C}}{0.15\text{m}}\\ =2.4\times {10}^5\text{V}\\ =\text{0.24MV}\end{array}$

Hence, the electric potential is $\text{0.24MV}$.