A point charge q is situated at a distance d | vedclass

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Question

(D) Consider a small element of length $dx$ on the rod at a distance $x$ from the point charge $q$. The charge on this element is $dq = \frac{Q}{L} dx

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Consider a small element of length $dx$ on the rod at a distance $x$ from the point charge $q$. The charge on this element is $dq = \frac{Q}{L} dx$.The small electric force $dF$ between the point charge $q$ and the element $dq$ is given by Coulomb's law:$dF = \frac{1}{4\pi \epsilon_0} \frac{q \cdot dq}{x^2} = \frac{1}{4\pi \epsilon_0} \frac{q (Q/L) dx}{x^2}$.To find the total force $F$,integrate $dF$ from $x = d$ to $x = d+L$:$F = \int_{d}^{d+L} \frac{qQ}{4\pi \epsilon_0 L} \frac{dx}{x^2} = \frac{qQ}{4\pi \epsilon_0 L} \left[ -\frac{1}{x} \right]_{d}^{d+L}$.$F = \frac{qQ}{4\pi \epsilon_0 L} \left( \frac{1}{d} - \frac{1}{d+L} \right) = \frac{qQ}{4\pi \epsilon_0 L} \left( \frac{d+L-d}{d(d+L)} \right)$.$F = \frac{qQ}{4\pi \epsilon_0 L} \left( \frac{L}{d(d+L)} \right) = \frac{1}{4\pi \epsilon_0} \frac{qQ}{d(d+L)}$.Thus,the correct option is $D$.

$A$ point charge $q$ is situated 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. Then the magnitude of the electric force between them is:

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