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Electric Potential At The Edge Of A Disc

Electric Potential At The Edge Of A Disc :- Let us calculate the electric potential at the edge of a disc of radius R, which is uniformly charged and has a surface charge density σ.

Let A be the point on the diameter AB at the edge of the disc at which we have to calculate electric potential. Taking A as centre, we draw two arcs of radii r and (r+dr) as shown in figure above. The area of the infinitely small region having thickness dr between the arcs is :

$\displaystyle dA=r(2\theta )dr=2\theta rdr$

where 2θ is the angle subtended by this area element PQ at the point A. Electric potential at point A due the element PQ is :

$\displaystyle dV=k\frac{dq}{r}=k\frac{\sigma dA}{r}=k\frac{\sigma 2\theta rdr}{r}=2k\sigma \theta dr$

In ΔAPB, we have

$\displaystyle r=2Rcos\theta$

$\displaystyle \Rightarrow dr=-2Rsin\theta d\theta$

Hence,

$\displaystyle dV=2k\sigma \theta (-2Rsin\theta d\theta )=-4Rk\sigma \theta sin\theta d\theta$

Total electric potential at point A due to whole disc,

$\displaystyle V=\int{dV=}-4Rk\sigma \int\limits_{\pi /2}^{0}{\theta sin\theta d\theta }$

Using integration by parts, i.e.,

$\displaystyle \int{(uv)dx}=u\int{vdx}-\int{\left\{ \frac{du}{dx}\int{vdx} \right\}dx}$

We get,

$\displaystyle V=-4Rk\sigma \left[ \theta \int{sin\theta d\theta }-\int{\left\{ \frac{d\theta }{d\theta }\int{sin\theta d\theta } \right\}d\theta } \right]_{\pi /2}^{0}$

$\displaystyle V=-4Rk\sigma \left[ \theta \left( -cos\theta \right)-\int{1.\left\{ -cos\theta \right\}d\theta } \right]_{\pi /2}^{0}$

$\displaystyle V=-4Rk\sigma \left[ -\theta cos\theta +\int{cos\theta d\theta } \right]_{\pi /2}^{0}$

$\displaystyle V=-4Rk\sigma \left[ -\theta cos\theta +sin\theta \right]_{\pi /2}^{0}$

$\displaystyle V=-4Rk\sigma \left[ \left( -0cos0+sin0 \right)-\left( -\frac{\pi }{2}cos\frac{\pi }{2}+sin\frac{\pi }{2} \right) \right]$

$\displaystyle V=-4Rk\sigma \left[ \left( -0+0 \right)-\left( -\frac{\pi }{2}\times 0+1 \right) \right]$

$\displaystyle V=-4Rk\sigma \left[ -1 \right]$

$\displaystyle \therefore V=4Rk\sigma =\frac{\sigma R}{\pi {{\varepsilon }_{0}}}$

Electric Potential At The Edge Of A Disc

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