Charge Q is uniformly distributed on a dielectric rod AB of length 2l. The potential at P shown in the figure is equal to

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$\frac{q}{4 {πϵ}_{0} l} \ln (2)$

$\frac{q}{4 {πϵ}_{0} 2 l}$

$\frac{q}{4 {πϵ}_{0} 2 l} \ln (3)$

$\frac{2 q}{4 {πϵ}_{0} l} \ln (2)$

answer is C.

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Detailed Solution

$V = \int dV = \int \frac{1}{4 {πε}_{0}} \frac{dq}{x}$
Question Image
$\begin{array}{l} = \frac{1}{4 {πε}_{0}} \int \frac{λdx}{x} = \frac{λ}{4 {πε}_{0}} \int \frac{dx}{x} = {\frac{λ}{4 {πε}_{0}} \log x|}_{l}^{3 l} \\ = \frac{λ}{4 {πε}_{0}} (\log 3 l - \log l) = \frac{λ}{4 {πε}_{0}} (\log 3) \\ = \frac{q}{4 {πε}_{0} 2 l} (\log 3) \end{array}$