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?
3 answers
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?
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 x-2m (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)/(2-x)^2
Since they are both in µ C, rewrite the equation as
(50/x^2)=( -20)/(2-x)^2)
50(2-x)^2=-20(x^2)
50(x^2-4x+4)=-20x^2
50x^2-200x+200=-20x^2
factoring out 10,
5x^2-20x+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.
Let the distance between Q1 and Qo equal x(r1), and the distance between Q2 and Qo equal x-2m (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)/(2-x)^2
Since they are both in µ C, rewrite the equation as
(50/x^2)=( -20)/(2-x)^2)
50(2-x)^2=-20(x^2)
50(x^2-4x+4)=-20x^2
50x^2-200x+200=-20x^2
factoring out 10,
5x^2-20x+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.
Last set of values should read:
7x^2-20x+20=0
7x^2-20x+20=0
2 answers