Question

Two charges 2 x 10-6 & 1 x 10-6 are placed at a separation of 10 cm. Where should a third charge be placed such that it experiences no net force due to these charges????? Please answer to this question!!!!!!. This question is very very important for me!!!!!!.

Answer 1

let '*' be the first charge which is 2×10^-6 and '+' be other charge which is 10^-6
*__________#_____________+
the distance between * and + is given 10cm, let the distance from * to # be x, so the distance between #and + will be (10-x) ( we have to take 10cm as 0.1 m )
formula for force between 2 charges is.
F= kqQ/r^2
as the net force between them is zero.
so ,F(*#)-F(#+) =0
F(*#)=F(#+)
where F(*#)= K×q(#)×2×10^-6/x^2
AND. F(#+)= K×q(#)×10^-6/(0.1-x) ^2
SO, K×q(#)×2×10^-6/x^2=K×q(#)×10^-6/(0.1-x)^2
2×10^-6/x^2=10^-6/(0.1-x)^2
2/x^2=1/(0.1-x) ^2
solve this equation and u will get our answer ook, u may take 0.1 m as 10 cm. u can replace it by 10 cm, it will be easy for further calculations
BY THE WAY IN WHICH STANDARD U R?