Three point charges q,-2q,and 2q are placed o | vedclass

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(B) Let the charges be $q_1 = q$ at $x=0$,$q_2 = -2q$ at $x=\frac{3}{4}R$,and $q_3 = 2q$ at $x=R$.The force on $q_2$ due to $q_1$ is $F_{21} = \frac{k

Vedclass · Accepted Answer

$5440$

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(B) Let the charges be $q_1 = q$ at $x=0$,$q_2 = -2q$ at $x=\frac{3}{4}R$,and $q_3 = 2q$ at $x=R$.The force on $q_2$ due to $q_1$ is $F_{21} = \frac{k |q_1 q_2|}{r_{21}^2} = \frac{k |q(-2q)|}{(\frac{3}{4}R)^2} = \frac{2kq^2}{\frac{9}{16}R^2} = \frac{32kq^2}{9R^2}$ (directed towards the origin,i.e.,in $-x$ direction).The force on $q_2$ due to $q_3$ is $F_{23} = \frac{k |q_2 q_3|}{r_{23}^2} = \frac{k |(-2q)(2q)|}{(R - \frac{3}{4}R)^2} = \frac{4kq^2}{(\frac{1}{4}R)^2} = \frac{4kq^2}{\frac{1}{16}R^2} = \frac{64kq^2}{R^2}$ (directed towards $x=R$,i.e.,in $+x$ direction).The net force on $-2q$ is $F_{net} = F_{23} - F_{21} = \frac{64kq^2}{R^2} - \frac{32kq^2}{9R^2} = \frac{576kq^2 - 32kq^2}{9R^2} = \frac{544kq^2}{9R^2}$.Substituting values $k = 9 	imes 10^9 \, N \cdot m^2/C^2$,$q = 2 	imes 10^{-6} \, C$,and $R = 0.02 \, m$:$F_{net} = \frac{544 	imes 9 	imes 10^9 	imes (2 	imes 10^{-6})^2}{9 	imes (0.02)^2} = \frac{544 	imes 10^9 	imes 4 	imes 10^{-12}}{4 	imes 10^{-4}} = 544 	imes 10^1 = 5440 \, N$.

Three point charges $q$,$-2q$,and $2q$ are placed on the $x$-axis at distances $x=0$,$x=\frac{3}{4}R$,and $x=R$ respectively from the origin as shown. If $q = 2 \times 10^{-6} \, C$ and $R = 2 \, cm$,the magnitude of the net force experienced by the charge $-2q$ is .......... $N$.

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