Discussion Forum

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

Answer:

Explanation:

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

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

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

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

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

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