The Brainliest Answer!

This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
See diagram.

Potential dV due to a small element dx of the rod AB at a distance x  from A.  The element dx has a charge = dq = Q/L * dx.

The potential (electrostatic) due to dq at point O, which is L distance away from A
     = dV=\frac{1}{4\pi\epsilon}\frac{dq}{\sqrt{L^2+x^2}}\\\\=\frac{Q}{4\pi\epsilon L^2}\frac{dx}{\sqrt{1+(\frac{x}{L})^2}}\\\\x=x'L\\\\dV=\frac{Q}{4\pi\epsilon L}\frac{dx'}{\sqrt{1+x'^2}}\\\\V= \frac{Q}{4\pi\epsilon L}\int\limits^1_0 {\frac{1}{\sqrt{1+x'^2}}} \, dx' \\\\= \frac{Q}{4\pi\epsilon L}[tan^{-1}\ x']_0^1\\\\=\frac{Q}{4\pi\epsilon L}\frac{\pi}{4}

2 5 2
excuse me pls. In the above answer, the integral answer is not tan^-1 x. It is Ln [ x' + sqroot(1+x'^2) ]. The upper limit is 1 and lower limit is 0. Evaluating this expression with these limits gives: Ln (1 + sqroot(2)).
Hence, the answer is Q/(4πεL) * Ln (1+√2). This is correct.
thankyou very much