Elementry Electrostatics question

[838]

Homework Statement

Two point charges (q1=+20 nC and q2=-40 nC) are placed 10.0 cm apart onlong the x-axis. Where can a third charge (q3=+20 nC) be placed so that the net electrostatic force on that charge is zero?

Homework Equations

F=k(q1q2)/r^2

The Attempt at a Solution

It's definitely not a hard problem, I'm guessing I'm supposed to set the entire equation into equilibrium. Other than that basic concept, I really don't understand how to solve for this problem. My professor only went over calculating forces between two charges.

Thanks for any help, I could really use it. :(

 

[Dick]

Place q1 at x=0 and q2 at x=10, and put q3 at x=a. Now write the two forces on q3 as a function of a, set them equal and solve for a. It's helpful when you do this to figure out what portion of the line you expect q3 to be on first. Will a<0, 0<a<10 or a>10?

 

[838]

So if I'm understanding you correctly, it should look like this:

k*(q1q3)/a^2=k*(q2q3)/(10+a)^2 since a < 0. I think...

So we get a = 4.14

To test this, I can calculate the force between q1 and q3 and match it with q2 and q3, since they should be equal.

So, k*(q1q3)/(4.14^2) = 23.31k
and k*(q2q3)/(10-4.14)^2=23.31k.

Should be correct. Any errors?

Thanks.

 

[Dick]

Careful! I think you are picking the wrong root. You want solve k*(q1q3)/a^2=k*(q2q3)/(10-a)^2 if a<0 is the coordinate of the point. If you solve that you get 4.14 and -24.14. Since you assumed a<0 the one to pick is -24.14. If you call a the distance of the point from the origin then you solve k*(q1q3)/a^2=k*(q2q3)/(10+a)^2 and assuming a>0. You get -4.14, 24.14 and pick the positive one.

 

[838]

Ah, ok, I understand. I wasn't sure which root to choose for. Thanks a lot!

 

[Dick]

Ah, ok, I understand. I wasn't sure which root to choose for. Thanks a lot!

The root you chose corresponds to the point between the charge at x=0 and the charge at x=10. Here the magnitude of the forces are equal, but they point in the same direction.