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2015-04-02T07:31:44+05:30

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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