When a $5 \mathrm{C}$ charge is kept in a uniform electric field, a force of $5000 \mathrm{~N}$ acts on it.…
When a $5 \mathrm{C}$ charge is kept in a uniform electric field, a force of $5000 \mathrm{~N}$ acts on it. Find the potential difference between two points in that field, separated by a distance of $1 \mathrm{~cm}$.
$10 \mathrm{~V}$
$250 \mathrm{~V}$
) $1000 \mathrm{~V}$
$2500 \mathrm{~V}$
Solution
Charge, $q=5 \mathrm{C}$
Force,
$
\begin{aligned}
& F=5000 \mathrm{~N} \\
& d=1 \mathrm{~cm}=10^{-2} \mathrm{~m}
\end{aligned}
$
$\therefore$ Potential difference between the given points,
$
\begin{aligned}
V & =B \times d=\frac{F}{q} \times d & {\left[\because B=\frac{F}{q}\right] } \\
& =\frac{5000}{5} \times 10^{-2}=10 \mathrm{~V} &
\end{aligned}
$