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(C) Consider a small element of the rod of length $dx$ at a distance $x$ from the point charge $q$,as shown in the figure. Treat this element as a poi

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

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(C) Consider a small element of the rod of length $dx$ at a distance $x$ from the point charge $q$,as shown in the figure. Treat this element as a point charge.The force $dF$ between the charge $q$ and the charge element $dQ$ is given by Coulomb's law:$dF = \frac{1}{4\pi \epsilon_0} \frac{q \cdot dQ}{x^2}$Since the charge $Q$ is uniformly distributed over length $L$,the charge per unit length is $\lambda = \frac{Q}{L}$. Therefore,$dQ = \lambda dx = \frac{Q}{L} dx$.Substituting $dQ$ into the force equation:$dF = \frac{1}{4\pi \epsilon_0} \frac{q (Q/L) dx}{x^2} = \frac{qQ}{4\pi \epsilon_0 L} \frac{dx}{x^2}$To find the total force $F$,integrate $dF$ from $x = d$ to $x = d + L$:$F = \int_{d}^{d+L} \frac{qQ}{4\pi \epsilon_0 L} \frac{dx}{x^2} = \frac{qQ}{4\pi \epsilon_0 L} \left[ -\frac{1}{x} \right]_{d}^{d+L}$$F = \frac{qQ}{4\pi \epsilon_0 L} \left( -\frac{1}{d+L} - (-\frac{1}{d}) \right) = \frac{qQ}{4\pi \epsilon_0 L} \left( \frac{1}{d} - \frac{1}{d+L} \right)$$F = \frac{qQ}{4\pi \epsilon_0 L} \left( \frac{(d+L) - d}{d(d+L)} \right) = \frac{qQ}{4\pi \epsilon_0 L} \left( \frac{L}{d(d+L)} \right)$$F = \frac{1}{4\pi \epsilon_0} \frac{qQ}{d(d+L)}$

$A$ point charge $q$ is placed at a distance $d$ from one end of a thin non-conducting rod of length $L$ having a charge $Q$ uniformly distributed along its length,as shown in the figure. Find the magnitude of the electric force between them.

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