Answer
$V \approx\dfrac{\pi k_e \sigma R^2}{d}$ for large $d$
Work Step by Step
Here, we have $\sqrt{d^2+R^2}=\sqrt{d^2(1+\dfrac{R^2}{d^2})}=d(1+\dfrac{R^2}{d^2})^{(1/2)}$
Use the Binomial series.
$d(1+\dfrac{R^2}{d^2})^{1/2}=d(1+\dfrac{1}{2}\dfrac{R^2}{d^2}+...)=d+\dfrac{1}{2d}(R^2)+....$
We are given that the expression for the potential is as follows:
$V \approx 2\pi k_e \sigma (d+\dfrac{R^2}{2d}+......-d)$
or, $V \approx 2\pi k_e \sigma (\dfrac{R^2}{2d})$
Hence, $V \approx\dfrac{\pi k_e \sigma R^2}{d}$ for large $d$
