Two equally charged small balls at a fixed distances experience a forceF. a similar uncharged ball, after touching one of them is placed at the middle point between the two balls. the force experienced by this ball is..??

Answers (1)

let the charges be q and q separated by a distance 2d apart. When an uncharged ball is touched with one of them then the charge gets distributed equally.

so the charge at the ends are q/2 and q and at the middle is q/2

initially, before the third charge is introduced force=F

F=\frac{Kq^2}{4d^2}

After the third charge is introduced

F_1=\frac{Kq\times(q/2)}{d^2}=\frac{Kq^2}{2d^2}

F_2=\frac{K(q/2)\times(q/2)}{d^2}=\frac{Kq^2}{4d^2}

The net force on the charge at the centre

\\F1-F2=\frac{Kq^2}{2d^2}-\frac{Kq^2}{4d^2}=\frac{Kq^2}{4d^2}=F

 

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