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(B) The work done $W$ in bringing a charge $q_3$ from infinity to a point is equal to the change in potential energy of the system,which is $W = q_3 \

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$4.5 	imes 10^{-3} \ J$

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(B) The work done $W$ in bringing a charge $q_3$ from infinity to a point is equal to the change in potential energy of the system,which is $W = q_3 	imes V$,where $V$ is the electric potential at the point $x = 2 \ m$ due to charges $q_1$ and $q_2$.The potential $V$ at $x = 2 \ m$ is given by:$V = \frac{1}{4\pi \epsilon_0} \left( \frac{q_1}{r_1} + \frac{q_2}{r_2} ight)$Here,$r_1$ is the distance from $q_1$ to $x = 2 \ m$,so $r_1 = 2 \ m$. $r_2$ is the distance from $q_2$ to $x = 2 \ m$,so $r_2 = 1 \ m$.Substituting the values:$W = q_3 	imes \frac{1}{4\pi \epsilon_0} \left( \frac{-1 	imes 10^{-6}}{2} + \frac{1 	imes 10^{-6}}{1} ight)$$W = (9 	imes 10^9) 	imes (1 	imes 10^{-6}) 	imes \left( -0.5 	imes 10^{-6} + 1 	imes 10^{-6} ight)$$W = 9 	imes 10^3 	imes (0.5 	imes 10^{-6})$$W = 4.5 	imes 10^{-3} \ J$

Two point charges are located on the $x$-axis. $q_1 = -1 \ \mu C$ at $x = 0$ and $q_2 = +1 \ \mu C$ at $x = 1 \ m$. Find the work done in bringing a third charge $q_3 = +1 \ \mu C$ from infinity to $x = 2 \ m$.

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