A charge Q is uniformly distributed over a long rod AB of length L .The point is placed left of point A at a distance L.Find electric potential at point O
Answer is Q ln(2)/4 pi absolute electric permeability L
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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}](https://tex.z-dn.net/?f=dV%3D%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon%7D%5Cfrac%7Bdq%7D%7B%5Csqrt%7BL%5E2%2Bx%5E2%7D%7D%5C%5C%5C%5C%3D%5Cfrac%7BQ%7D%7B4%5Cpi%5Cepsilon+L%5E2%7D%5Cfrac%7Bdx%7D%7B%5Csqrt%7B1%2B%28%5Cfrac%7Bx%7D%7BL%7D%29%5E2%7D%7D%5C%5C%5C%5Cx%3Dx%27L%5C%5C%5C%5CdV%3D%5Cfrac%7BQ%7D%7B4%5Cpi%5Cepsilon+L%7D%5Cfrac%7Bdx%27%7D%7B%5Csqrt%7B1%2Bx%27%5E2%7D%7D%5C%5C%5C%5CV%3D+%5Cfrac%7BQ%7D%7B4%5Cpi%5Cepsilon+L%7D%5Cint%5Climits%5E1_0+%7B%5Cfrac%7B1%7D%7B%5Csqrt%7B1%2Bx%27%5E2%7D%7D%7D+%5C%2C+dx%27+%5C%5C%5C%5C%3D+%5Cfrac%7BQ%7D%7B4%5Cpi%5Cepsilon+L%7D%5Btan%5E%7B-1%7D%5C+x%27%5D_0%5E1%5C%5C%5C%5C%3D%5Cfrac%7BQ%7D%7B4%5Cpi%5Cepsilon+L%7D%5Cfrac%7B%5Cpi%7D%7B4%7D "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}")
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
=
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kvnmurty:
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
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