Question:
Calculate potential on the axis of a disc of radius R due to a charge Q uniformly distributed on its surface.
Solution:
In the above figure, we can see that the disc is divided into several charged rings. Let P be the point on the axis of the disc at a distance x from the centre of the disc.
The radius of the ring is r and the width is dr. dq is the charge on the ring which is given as
dq = σdA = σ2πrdr
The potential is given as
$d V=\frac{1}{4 \pi \epsilon_{0}} \frac{d q}{\sqrt{r^{2}+x^{2}}}=\frac{1}{4 \pi \epsilon_{0}} \frac{(\sigma 2 \pi r d r)}{\sqrt{r^{2}+x^{2}}}$
The total potential at P is given as
Q/2πε0R2 (√R2 + x2 –x)