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)
S safeer

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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