A charge Q is uniformly distributed over a lo | vedclass

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Question

(A) Let the linear charge density of the rod be $\lambda = \frac{Q}{L}$.Consider a small element of length $dx$ on the rod at a distance $x$ from poin

Vedclass · Accepted Answer

$\frac{Q \ln 2}{4 \pi \varepsilon_0 L}$

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(A) Let the linear charge density of the rod be $\lambda = \frac{Q}{L}$.Consider a small element of length $dx$ on the rod at a distance $x$ from point $O$.The charge on this element is $dq = \lambda dx = \frac{Q}{L} dx$.The electric potential $dV$ at point $O$ due to this element is given by $dV = \frac{1}{4 \pi \varepsilon_0} \frac{dq}{x} = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{L} \frac{dx}{x}$.The total potential $V$ at point $O$ is obtained by integrating from $x = L$ (at end $A$) to $x = 2L$ (at end $B$):$V = \int_{L}^{2L} \frac{Q}{4 \pi \varepsilon_0 L} \frac{dx}{x} = \frac{Q}{4 \pi \varepsilon_0 L} [\ln x]_{L}^{2L} = \frac{Q}{4 \pi \varepsilon_0 L} (\ln 2L - \ln L) = \frac{Q \ln 2}{4 \pi \varepsilon_0 L}$.

$A$ 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 a distance $L$ from the end $A$ is

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