The easy answer is at infinity.
let x be the distance from origin.
F5=F7
kq1 q/x^2= kq2 q/(1-x)^2
note k divides out, q divides out, so the charge on q does not matter.
5E-6 *(1-2x+x^2)= 7E-6*x^2
5-10x+5x^2)-7x^2=0
2x^2+10x-5=0
x=(-10+-sqrt(100+40)/4
x=2.5+-1/4 11.8
x=2.5+-2.95=.45m for one position
x=(10+-sqrt(100+40)/4
x=2.5+-.5*sqrt(140)
=2.5+-.5*11.8
If one charge (+5 uC) is placed at origin and a second charge of (+7 uC) at x = 100 cm, then where can a third charge be placed and of what size, so that it has no electrostatic force on it.
I know the magnitude of the force will not matter, however, I'm not sure as to how to determine the distance.
Cheers.
1 answer
1 answer