Question

A charge of $5\text{C}$ experiences a force of $5000\text{N}$ when it is kept in a uniform electric field. Find the potential difference (in$\text{V}$) between two points seperated by a distance of $1\text{cm}$.

• A. 10
• B. 10.0
• C. 10.00

Answer 1

Answer: (A), (B), (C)

Given,

Charge, $q=5\text{C}$
Force experienced by charge, $F=5000\text{N}$
Seperation between given two points for which the potential difference is calculated is $d=1\text{cm}$

When a charge is kept in an electric field $E$, Force acting on the charge is given by
$F=qE$
We know from the relation between uniform electric field $E$ and potential $V.$
$E=\frac{V}{d}$

$\therefore F=q\left(\frac{V}{d}\right)$
Substituting the given data we get,
$5000=\frac{5\times V}{{10}^{-2}}$

$\Rightarrow V=10\text{volt}$