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(B) Let the linear charge density be $\lambda = Q/L$. Consider a small element $dx$ at a distance $x$ from the charge $q$. The force $dF$ on charge $q

Vedclass · Accepted Answer

$\frac{qQ}{4\pi \epsilon_0 (r^2 - (L/2)^2)}$

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(B) Let the linear charge density be $\lambda = Q/L$. Consider a small element $dx$ at a distance $x$ from the charge $q$. The force $dF$ on charge $q$ due to this element is $dF = \frac{1}{4\pi \epsilon_0} \frac{q (\lambda dx)}{x^2}$.Integrating this from $x = r - L/2$ to $x = r + L/2$:$F = \int_{r - L/2}^{r + L/2} \frac{q \lambda}{4\pi \epsilon_0 x^2} dx$$F = \frac{q \lambda}{4\pi \epsilon_0} \left[ -\frac{1}{x} \right]_{r - L/2}^{r + L/2}$$F = \frac{q (Q/L)}{4\pi \epsilon_0} \left( \frac{1}{r - L/2} - \frac{1}{r + L/2} \right)$$F = \frac{qQ}{4\pi \epsilon_0 L} \left( \frac{(r + L/2) - (r - L/2)}{r^2 - (L/2)^2} \right)$$F = \frac{qQ}{4\pi \epsilon_0 L} \left( \frac{L}{r^2 - (L/2)^2} \right)$$F = \frac{qQ}{4\pi \epsilon_0 (r^2 - (L/2)^2)}$

$A$ charge $Q$ is distributed over a line of length $L$. Another point charge $q$ is placed at a distance $r$ from the centre of the line distribution. Then the force experienced by $q$ is

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