- Home
- Standard 12
- Physics
Calculate potential on the axis of a ring due to charge $Q$ uniformly distributed along the ring of radius $R$.
Solution
Suppose, $+Q$ charge uniformly distributed on the ring of radius $R=a$
Let us take point P to be at a distance $x$ from the centre of the ring. The charge $d q$ is at a distance $r$ from the point P then,
$r=\sqrt{x^{2}+a^{2}}$
and potential at $\mathrm{P}$ due to $d q \mathrm{~V}=\frac{k d q}{r}$
Potential at P due to charge on the whole ring,
$\mathrm{V}=k \int \frac{d q}{r}=k \int \frac{d q}{\sqrt{x^{2}+a^{2}}}$
$\mathrm{~V}=\frac{k}{\sqrt{x^{2}+a^{2}}} \int d q=\frac{k \mathrm{Q}}{\sqrt{x^{2}+a^{2}}}\left[\because \int d q=\mathrm{Q}\right]$
$\therefore$ The net electric potential,
$\mathrm{V}=\frac{\mathrm{Q}}{4 \pi \in_{0} \sqrt{x^{2}+a^{2}}}$
