Question

At a certain distance from a point charge, the field intensity is $500\text{V/m}$ and the electric potential is $-3000\text{V}$. Then the distance from charge is

• A. $6\text{m}$
• B. $4\text{m}$
• C. $8\text{m}$
• D. $2\text{m}$

Answer 1

The correct option is A $6\text{m}$

We know, electric field at a point due to a point charge $q$ is $E=\frac{kq}{r^2}$ and electric potential is $V=\frac{kq}{r}$
where $r$ is the distance of point from the charge.
$\Rightarrow E=\frac{kq}{r}.\frac{1}{r}$
$\text{Or,}E=\frac{-V}{r}$

[Negative sign has been included because it shows that, as we move along $\overrightarrow{E}$, the electric potential$\left(V\right)$ decreases]


$\Rightarrow r=\frac{-V}{E}$
$\Rightarrow r=\frac{(-3000)}{500}$
$\therefore r=6\text{m}$

$\begin{array}{c}\text{Why this question?}\\ \text{concept - In electric field of a point charge, as we move along}\overrightarrow{E},\\ \text{electric potential decreases}.\\ \text{Here},V_A>V_B\text{as}{r}_A<r_B\end{array}$