Question

Equipotential spheres are drawn around a point charge $+q$. As we move away from the charge, the spacing between two spheres having a constant potential difference

• A. Increases
• B. Decreases
• C. Remains constant
• D. First decrease then increases

Answer 1

The correct option is A Increases


Potential due to point charge at distance $r$ is given by

$V=\frac{kq}{r}$

Thus, $V_1=\frac{kq}{r_1}=\frac{q}{4\pi {ϵ}_0{r}_1}$

$V_2=\frac{kq}{r_2}=\frac{q}{4\pi {ϵ}_0{r}_2}$

$\because r_2>r_1\text{hence}{V}_1>V_2$

$\Rightarrow V_1-V_2=\frac{q}{4\pi {ϵ}_0}(\frac{1}{r_1}-\frac{1}{r_2})=\frac{q}{4\pi {ϵ}_0}\left(\frac{r_2-r_1}{r_1{r}_2}\right)$

Spacing between two spheres,

$r_2-r_1=\frac{4\pi {ϵ}_0(V_1-V_2)}{q}\left(r_1{r}_2\right)$

For constant potential difference between spheres

$V_1-V_2=\text{constant}$

$\Rightarrow r_2-r_1\propto r_1{r}_2$

As we move away from charge the product $r_1{r}_2$ increase.

Hence, $(r_2-r_1)\uparrow$

$\therefore$ Spacing between two spheres increases.

Hence, option $\left(a\right)$ is the correct answer.

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: A point charge is a source of non-uniform}\overrightarrow{E}.\\ E=\frac{kq}{r^2}\\ \text{Thus, uniformly spaced equipotential surface i.e}\\ \text{sphere will never exist due to point charge}.\end{array}$