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(D) The linear charge density of the rod is $\lambda = \frac{Q}{L}$.Consider a small element of length $dx$ at a distance $x$ from point $O$.The charg

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$\frac{Q \ln 2}{4\pi \epsilon_0 L}$

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(D) The linear charge density of the rod is $\lambda = \frac{Q}{L}$.Consider a small element of length $dx$ at a distance $x$ from point $O$.The charge of this element is $dq = \lambda dx = \frac{Q}{L} dx$.The potential $dV$ at point $O$ due to this element is $dV = \frac{k dq}{x} = \frac{1}{4\pi \epsilon_0} \frac{Q}{L} \frac{dx}{x}$.To find the total potential $V$,integrate from $x = L$ to $x = 2L$:$V = \int_{L}^{2L} \frac{Q}{4\pi \epsilon_0 L} \frac{dx}{x} = \frac{Q}{4\pi \epsilon_0 L} [\ln x]_{L}^{2L}$.$V = \frac{Q}{4\pi \epsilon_0 L} (\ln 2L - \ln L) = \frac{Q}{4\pi \epsilon_0 L} \ln\left(\frac{2L}{L}\right) = \frac{Q \ln 2}{4\pi \epsilon_0 L}$.

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

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