A point charge q = 1 mu C is located at a dis | vedclass

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(B) Let the point charge $q$ be at the origin $x=0$. The wire extends from $x = 2 	ext{ cm}$ to $x = 12 	ext{ cm}$.The linear charge density of the

Vedclass · Accepted Answer

$90$

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(B) Let the point charge $q$ be at the origin $x=0$. The wire extends from $x = 2 	ext{ cm}$ to $x = 12 	ext{ cm}$.The linear charge density of the wire is $\lambda = \frac{Q}{L} = \frac{24 	imes 10^{-6} 	ext{ C}}{0.1 	ext{ m}} = 2.4 	imes 10^{-4} 	ext{ C/m}$.Consider a small element $dx$ of the wire at a distance $x$ from the point charge $q$. The charge on this element is $dq = \lambda dx$.The force between the point charge $q$ and the element $dq$ is $dF = \frac{k q dq}{x^2} = \frac{k q \lambda dx}{x^2}$, where $k = 9 	imes 10^9 	ext{ N} \cdot 	ext{m}^2 / 	ext{C}^2$.The total force $F$ is the integral of $dF$ from $x = 2 	ext{ cm} = 0.02 	ext{ m}$ to $x = 12 	ext{ cm} = 0.12 	ext{ m}$:$F = \int_{0.02}^{0.12} \frac{k q \lambda}{x^2} dx = k q \lambda \left[ -\frac{1}{x} ight]_{0.02}^{0.12} = k q \lambda \left( \frac{1}{0.02} - \frac{1}{0.12} ight)$$F = (9 	imes 10^9) 	imes (1 	imes 10^{-6}) 	imes (2.4 	imes 10^{-4}) 	imes \left( 50 - 8.333 ight) = 9000 	imes 2.4 	imes 10^{-4} 	imes \left( \frac{6-1}{0.12} ight) = 9000 	imes 2.4 	imes 10^{-4} 	imes \frac{5}{0.12} = 90 	ext{ N}$.

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