Potential of the rod on its axis.

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SUMMARY

The potential at point P on the axis of a uniformly charged rod of length l and charge q is calculated using the formula: \(\varphi(r) = \frac{1}{4\pi\varepsilon_0\varepsilon}\frac{q}{l}\ln {\frac{r+l}{r}}\). This formula results from integrating the potential contributions from each infinitesimal segment of the rod. The distance from the rod to point P is denoted as r, and the integration limits are from r to r+l.

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Sumedh
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If a rod of length l is uniformly charged with charge q what will be the potential at point p?
The point is on the axis of the rod.Distance between the rod and point is r.
Also tell the concept behind this. Any help will be highly appreciated.
 

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Easy, just sum (integrate) potential coming from every point of your rod:
\varphi(r) = \int_r^{r+l}\frac{1}{4\pi\varepsilon_0\varepsilon}\frac{q}{l}\frac{1}{\varrho}d\varrho=<br /> \frac{1}{4\pi\varepsilon_0\varepsilon}\frac{q}{l}\ln {\frac{r+l}{r}}
 
thank you very much
 

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