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Calculate potential on the axis of a disc of radius R due to a charge Q uniformly distributed on its surface.

Answers (1)

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=\sigma dA=\sigma 2\pi rdr

The potential is given as

dV=\frac{1}{4\pi\varepsilon _{0}}\frac{dq}{\sqrt{r^{2}+x^{2}}}=\frac{1}{4\pi \varepsilon _{0}}\frac{\left ( \sigma 2\pi rdr \right )}{\sqrt{r^{2}+x^{2}}}

The total potential at P is given as

\frac{Q}{2\pi \varepsilon _{0}R^2}\left ( \sqrt{R_{2}+x_{2}-x} \right )

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