Posted by **Steve** on Wednesday, March 27, 2013 at 11:34am.

Two positive point charges are placed on the x-axis. One, of magnitude 4Q, is placed at the origin. The other, of magnitude Q is placed at x=3 m. Neither charge is able to move. Where on the x-axis in meters can I place a third positive point charge such that the magnitude of the net force on the third charge is zero?

- Physics -
**exactly**, Monday, October 7, 2013 at 1:34am
Lets assume a charge of q is placed at a point (x,o) where 0<x<3.

Your net force on that middle charge will be the sum of two electrostatic forces. Your plan is to use a distance of x for the force calculation between the leftmost charges and a distance of 3-x for the force calculation between the rightmost charges. These two forces should be equal to achieve equilibrium.

Basically, thanks to Coulomb's Law, I have 4Q/(x2) = Q/((3-x)2) after simplification. This yields

x^2-8x+12=0

(x-2)(x-6)=0

x=2 or x=6

Now just note that x has to be between 0 and 3 meters because that's the only way the field directions will oppose each other.

Anyways, this is quite a nice application of Coulomb's law. However, next time please refrain from posting live brilliant problems.

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