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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 the

Vedclass · Accepted Answer

$\frac{1}{4\pi \epsilon_0} \frac{qQ}{d(d+L)}$

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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 the point charge $q$.The charge of this element is $dq = \lambda dx = \frac{Q}{L} dx$.The electrostatic force $dF$ between this element and the point charge $q$ is given by Coulomb's law:$dF = \frac{1}{4\pi \epsilon_0} \frac{q dq}{x^2} = \frac{1}{4\pi \epsilon_0} \frac{q (Q/L) dx}{x^2}$.To find the total force $F$,we integrate $dF$ from $x = d$ to $x = d + L$:$F = \int_{d}^{d+L} \frac{1}{4\pi \epsilon_0} \frac{qQ}{L} \frac{dx}{x^2} = \frac{qQ}{4\pi \epsilon_0 L} \int_{d}^{d+L} x^{-2} dx$.$F = \frac{qQ}{4\pi \epsilon_0 L} \left[ -\frac{1}{x} \right]_{d}^{d+L} = \frac{qQ}{4\pi \epsilon_0 L} \left( -\frac{1}{d+L} - (-\frac{1}{d}) \right)$.$F = \frac{qQ}{4\pi \epsilon_0 L} \left( \frac{1}{d} - \frac{1}{d+L} \right) = \frac{qQ}{4\pi \epsilon_0 L} \left( \frac{d+L-d}{d(d+L)} \right)$.$F = \frac{qQ}{4\pi \epsilon_0 L} \left( \frac{L}{d(d+L)} \right) = \frac{1}{4\pi \epsilon_0} \frac{qQ}{d(d+L)}$.

The electrostatic force of interaction between a uniformly charged rod having total charge $Q$ and length $L$ and a point charge $q$ as shown in the figure is:

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