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Discussion Forum



1)

A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying  at distance L from the end A is

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A) $\frac{Q}{8\pi\epsilon_{0}L}$

B) $\frac{3Q}{4\pi\epsilon_{0}L}$

C) $\frac{Q}{4\pi\epsilon_{0}L l n 2}$

D) $\frac{Q ln 2}{4\pi\epsilon_{0}L}$

Answer:

Option D

Explanation:

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$V=\int_{L}^{2L} \frac{kdQ}{x}$

$=\int_{L}^{2L} \frac{k\left(\frac{Q}{L}\right)dx}{x}$

$=\frac{Q}{4 \pi \epsilon_{0}L}\int_{L}^{2L} \left(\frac{1}{x}\right)dx$

$=\frac{Q}{4 \pi \epsilon_{0}L}[\log_{e} x]_L^{2L}$

$=\frac{Q}{4 \pi \epsilon_{0}L}[\log_{e} 2L-\log_{e} L]$

$=\frac{Q}{4 \pi \epsilon_{0}L}ln(2)$