Answer
$V \approx\dfrac{\pi k_e \sigma R^2}{d}$ for large $d$
Work Step by Step
Consider $\sqrt{d^2+R^2}=\sqrt{d^2(1+\dfrac{R^2}{d^2})}=d(1+\dfrac{R^2}{d^2})^{1/2}$
Now, need to apply the Binomial series.
$d(1+\dfrac{R^2}{d^2})^{1/2}=d(1+\dfrac{R^2}{2d^2}+...)=d+\dfrac{R^2}{2d}+....$
The expression for potential is given as below:
$V \approx 2\pi k_e \sigma (d+\dfrac{R^2}{2d}+......-d)$
This can be further rewritten as below:
$V \approx 2\pi k_e \sigma (\dfrac{R^2}{2d})$
Hence, we get $V \approx\dfrac{\pi k_e \sigma R^2}{d}$ for large $d$
