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(D) The electric field due to an infinite line charge at a distance $r$ is given by $E = \frac{2k\lambda}{r}$.Here,$r_1 = 10 \, mm = 10^{-2} \, m$ and

Vedclass · Accepted Answer

$4.44 \, \mu C$

Vedclass · Answer

(D) The electric field due to an infinite line charge at a distance $r$ is given by $E = \frac{2k\lambda}{r}$.Here,$r_1 = 10 \, mm = 10^{-2} \, m$ and $r_2 = 12 \, mm = 12 	imes 10^{-3} \, m$.The force on the positive charge is $F_1 = qE_1 = q \left( \frac{2k\lambda}{r_1} ight)$ (directed away from the line charge).The force on the negative charge is $F_2 = qE_2 = q \left( \frac{2k\lambda}{r_2} ight)$ (directed towards the line charge).The net force is $F_{net} = F_1 - F_2 = 2k\lambda q \left( \frac{1}{r_1} - \frac{1}{r_2} ight)$.Substituting the values: $4 = 2 	imes (9 	imes 10^9) 	imes (3.0 	imes 10^{-6}) 	imes q 	imes \left( \frac{1}{10 	imes 10^{-3}} - \frac{1}{12 	imes 10^{-3}} ight)$.$4 = 54 	imes 10^3 	imes q 	imes \left( 100 - 83.33 ight) = 54 	imes 10^3 	imes q 	imes (16.67)$.$4 = 900180 	imes q \Rightarrow q \approx 4.44 	imes 10^{-6} \, C = 4.44 \, \mu C$.

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