When a charge of $3 \mathrm{C}$ is placed in uniform electric field, it experiences a force of $3000…
When a charge of $3 \mathrm{C}$ is placed in uniform electric field, it experiences a force of $3000 \mathrm{~N}$. Within this field, potential difference between two points separated by a distance of $1 \mathrm{~cm}$ is
$10 \mathrm{~V}$
$90 \mathrm{~V}$
$1000 \mathrm{~V}$
$3000 \mathrm{~V}$
Solution
Electric force, $F=q E$ and potential difference, $\mathrm{V}=\mathrm{Ed}$
$\therefore \quad \mathrm{V}=\frac{\mathrm{Fd}}{\mathrm{q}}=\frac{3000 \times 10^{-2}}{3}=10 \mathrm{~V}$