Potential with an edge

Consider a small circular arc of radius $x$ and thickness $dx$ centered at point $P$ on the edge as shown:

Clearly, applying cosine rule in $\triangle OPQ$ ,

$\cos \theta = \dfrac{x^2+R^2-R^2}{2RX} = \dfrac{x}{2R}$

Charge on the arc = $\sigma \times 2 \theta x dx$

Hence, the potential on the point $P$ due to this element = $dV = \dfrac{2 \sigma \theta x dx}{4 \pi \epsilon_{0} x}$

= $\dfrac{\sigma}{2 \pi \epsilon_{0}} \theta dx$

Hence, $V_{P} = \dfrac{\sigma}{2 \pi \epsilon_{0}} \displaystyle \int_{0}^{2R} \theta dx$

$\Rightarrow V_{P} = \dfrac{\sigma}{2 \pi \epsilon_{0}} \displaystyle \int_{0}^{2R} \cos^{-1} \bigg(\dfrac{x}{2R}\bigg) dx$

Putting $\frac{x}{2R}=t$ ,

$V_{P} = \dfrac{R \sigma}{\pi \epsilon_{0}} \displaystyle \int_{0}^{1} \cos^{-1} t dt = \boxed{\dfrac{R \sigma}{\pi \epsilon_{0}} }$

Put values to get answer as $18$

This problem is quite counter-intuitive, and although I still feel a bit skeptic about this one, the math just works out.

So here's an illustration of the problem. We want to find the potential at an arbitrary edge of the disc.

pic1

The goal is clearly to use calculus - integrate all of the differential potentials caused by the differential charges, but what function can model such a thing? $y = 2Rsin(x)$ can.

Imgur

Now we have set up everything necessary.

$V = \iint \frac{1}{4\pi \epsilon_o}\frac{dQ}{r} dA$

$= \frac{1}{4\pi \epsilon_o} \int\limits_0^\pi \int\limits_0^{2Rsin(\theta)} \frac{\sigma r}{r} dr d\theta = \large \frac{1}{4\pi \epsilon_o} \int\limits_0^\pi \int\limits_0^{2Rsin(\theta)} \sigma dr d\theta$

$= 4R\frac{\sigma}{4\pi \epsilon_o} = \boxed{18 }$ (volts)