Consider a particle with charge$\mathit{q}\mathbf{=}\mathbf{1}\mathbf{.50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{\text{C}}$ , and take$\mathit{V}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$ at infinity.

(a) What are the shape and dimensions of an equipotential surface having a potential of$\mathbf{\text{30.0V}}$ due to q alone?

(b) Are surfaces whose potentials differ by a constant amount ( $\mathbf{\text{1.0V}}$, say) evenly spaced?

The electric charge,$q=1.50\times {\text{10}}^{\text{-8}}\text{C}$, As$V=0$at infinity,

The electric potential,$V=30.0\text{V}$

Let the shape of the body be a sphere.

The equipotential surface should be spherical with radius R.

If the points in an electric field are all at the same electric potential, then they are known as the equipotential points.

Formula:

The electric potential is define by,

$V=\frac{1}{4\pi {\epsilon }_o}\cdot \frac{q}{R}$

Here V is electric potential, R is the distance between the point charges, q is charge, and $\frac{1}{4\pi {\epsilon }_o}$ is the constant having a value $9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$.

Electric Potential due to the spherical surface:

$\begin{array}{l}V=\frac{1}{4\pi {\epsilon }_o}\cdot \frac{q}{R}\\ 30.0=9.0\times {10}^9\times \frac{1.50\times {\text{10}}^{\text{-8}}}{R}\\ \begin{array}{c}R=\frac{9.0\times {10}^9\times 1.50\times {10}^{-8}}{30}\\ =\text{4.5m}\end{array}\end{array}$

Hence, electric potential due to spherical surface is $\text{4.5m}$.

If the electric field, $V=1.0\text{V}$

$\begin{array}{l}\frac{1}{4\pi {\epsilon }_o}\cdot \frac{q}{R}=1.0\\ 9.0\times {10}^9\times \frac{q}{R}=1.0\end{array}$

$\begin{array}{c}R=\frac{9.0\times {10}^9\times 1.50\times {10}^{-8}}{1.0}\\ =\text{135m}\end{array}$

Since,$R=135\text{m}$ is very large,so it is not possible to determine.