- Home
- Standard 12
- Physics
Calculate potential on the axis of a disc of radius $R$ due to a charge $Q$ uniformly distributed on its surface.
Solution
Let us consider a point $\mathrm{P}$ on the axis of the disc at a distance $x$ from the centre of the disc. Let the disc is divided into a numerous charged rings as shown in figure.
Let radius of ring $r$, width $d r$ and charges $d q$,
$\therefore \sigma d \mathrm{~A}=\sigma 2 \pi r d r$$..(1)$
Potential at $\mathrm{P}$,
$d \mathrm{~V}=\frac{k d q}{r}$
Charge on the ring $d q=+\sigma\left[\pi(r+d r)^{2}-\pi r^{2}\right]$
$\therefore d q=+\sigma \pi\left[(r+d r)^{2}-r^{2}\right]$
$\quad=+\sigma \pi\left[r^{2}+2 r d r+d r^{2}-r^{2}\right]$
$\quad=+\sigma \pi\left[2 r d r+d r^{2}\right]$
Neglecting $d r^{2}$ as $d r$ is very small,
$d q=2 \pi r \sigma d r$
and $d \mathrm{~V}=\frac{k d q}{\sqrt{r^{2}+x^{2}}}$$….(2)$
