Two point charges A and B of magnitude +8 tim | vedclass

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(B) Let the charges be $q_A = +8 	imes 10^{-6}\,C$ and $q_B = -8 	imes 10^{-6}\,C$. The distance between them is $d$. The midpoint $O$ is at a dista

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$3$

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(B) Let the charges be $q_A = +8 	imes 10^{-6}\,C$ and $q_B = -8 	imes 10^{-6}\,C$. The distance between them is $d$. The midpoint $O$ is at a distance $r = d/2$ from each charge.The electric field due to charge $A$ at $O$ is $E_A = \frac{K|q_A|}{r^2} = \frac{Kq}{(d/2)^2}$,directed towards $B$.The electric field due to charge $B$ at $O$ is $E_B = \frac{K|q_B|}{r^2} = \frac{Kq}{(d/2)^2}$,also directed towards $B$.The total electric field at $O$ is $E_0 = E_A + E_B = 2 	imes \frac{Kq}{(d/2)^2} = \frac{8Kq}{d^2}$.Given $E_0 = 6.4 	imes 10^4\,NC^{-1}$,$K = 9 	imes 10^9\,Nm^2C^{-2}$,and $q = 8 	imes 10^{-6}\,C$:$6.4 	imes 10^4 = \frac{8 	imes (9 	imes 10^9) 	imes (8 	imes 10^{-6})}{d^2}$$d^2 = \frac{8 	imes 9 	imes 10^9 	imes 8 	imes 10^{-6}}{6.4 	imes 10^4} = \frac{576 	imes 10^3}{6.4 	imes 10^4} = \frac{576}{64} = 9$$d = \sqrt{9} = 3\,m$.

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