Question

The electric potential due to a uniform disc of surface charge density $0.5C/m^2$ and radius 3 cm, at a point 4 cm from the centre on its perpendicular axis is

• A. $4{\epsilon }_0$
• B. $8{\epsilon }_0$
• C. $\frac{1}{4{\epsilon }_0}$
• D. $\frac{1}{8{\epsilon }_0}$

Answer 1

The correct option is C $\frac{1}{4{\epsilon }_0}$

So the formula to find the potential at an axial point for a charged disk is,
$V=\frac{2\pi \sigma [\sqrt{r^2+R^2}-r]}{4\pi {\epsilon }_0}$
Where,$\sigma$ is the charge density, R is the radius of the disk and r is the distance of the point from the center. If you don’t know where this formula came from, I encourage you to take a pen and a paper and derive it. It's very simple. Try it! Anyways, substituting in the values of corresponding variables we get,
$V=\frac{2\pi \times 0.5[\sqrt{3^2+4^2}-4]}{4\pi {\epsilon }_0}V=\frac{1}{4{\epsilon }_0}$