Posted by Lindsey on Monday, February 4, 2013 at 9:22pm.
Point charges Q1 = 50 µC and Q2 = 20 µC are placed 2 meters apart. Where can I put a third charge so that the net electrical force on it is zero?

Physics/Charges  Devron, Tuesday, February 5, 2013 at 7:20am
I'm a little rusty on this, but I think you need the charge of the one that you are putting in the middle. Was that provided to you?

Physics/Charges  Devron, Tuesday, February 5, 2013 at 7:59am
I apologize, you do not have to know it, it is not needed.
Let the distance between Q1 and Qo equal x(r1), and the distance between Q2 and Qo equal x2m (r2).
Using the sum of forces,
F1onO=F2onO
Ko[(Q1Qo)]/r^2=Ko[(Q1Qo)/r^2]
Ko and Qo cancel each other out leaving
Q1/r^2=Q2/r^2
Plugging in my values for Q1 and r, and r2
(50 x10^6 C/x^2)=( 20 x10^6 C)/(2x)^2
Since they are both in µ C, rewrite the equation as
(50/x^2)=( 20)/(2x)^2)
50(2x)^2=20(x^2)
50(x^24x+4)=20x^2
50x^2200x+200=20x^2
factoring out 10,
5x^220x+20=2x^2
7x^20x+20=0
Use the quadratic equation to solve for x
I hope this helps, but I am not that positive.

Physics/Charges  Devron, Tuesday, February 5, 2013 at 8:38am
Last set of values should read:
7x^220x+20=0
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