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

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Question

$F _{ BA }=\frac{ Kq (2 q )}{\left(\frac{3}{4} R \right)^2}=\frac{32 Kq ^2}{9 R ^2}$

$F _{ BC }=\frac{ K (2 q )(2 q )}{\left(\frac{ R }{4}\right)^2

Vedclass · Accepted Answer

$5440$

Vedclass · Answer

$F _{ BA }=\frac{ Kq (2 q )}{\left(\frac{3}{4} R ight)^2}=\frac{32 Kq ^2}{9 R ^2}$

$F _{ BC }=\frac{ K (2 q )(2 q )}{\left(\frac{ R }{4}ight)^2}=\frac{64 Kq ^2}{ R ^2}$

$F _{ B }= F _{ BC }- F _{ BA }=\frac{544 Kq ^2}{9 R ^2}$

$=\frac{544 	imes 9 	imes 10^9 	imes\left(2 	imes 10^{-6}ight)^2}{9 	imes\left(2 	imes 10^{-2}ight)^2}=5440\,N$

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

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