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(D) As shown in the figure,consider an elementary charge $dq$ having length $dx$ at a distance $x$ from the point charge $q$. The linear charge densit

Vedclass · Accepted Answer

$\frac{k Q q}{r(r+ L )}$

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(D) As shown in the figure,consider an elementary charge $dq$ having length $dx$ at a distance $x$ from the point charge $q$. The linear charge density is $\lambda = \frac{Q}{L}$.Then,$dq = \lambda dx = \frac{Q}{L} dx$.By Coulomb's law,the force $dF$ between the elementary charge $dq$ and the point charge $q$ is:$dF = \frac{k q dq}{x^2} = \frac{k q Q dx}{L x^2}$.To find the total force $F$,integrate $dF$ from $x = r$ to $x = r + L$:$F = \int_r^{r+L} \frac{k Q q}{L x^2} dx = \frac{k Q q}{L} \int_r^{r+L} x^{-2} dx$.$F = \frac{k Q q}{L} \left[ -\frac{1}{x} \right]_r^{r+L} = \frac{k Q q}{L} \left( -\frac{1}{r+L} - (-\frac{1}{r}) \right)$.$F = \frac{k Q q}{L} \left( \frac{1}{r} - \frac{1}{r+L} \right) = \frac{k Q q}{L} \left( \frac{r+L-r}{r(r+L)} \right)$.$F = \frac{k Q q}{L} \left( \frac{L}{r(r+L)} \right) = \frac{k Q q}{r(r+L)}$.

$A$ point charge $q$ is situated at a distance $r$ on the axis from one end of a thin conducting rod of length $L$ having a charge $Q$ (uniformly distributed along its length). The magnitude of the electric force between the two is . . . . . . .

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