Electrostatics: Two charged balls with attraction

AI Thread Summary
In the discussion about electrostatics, a user seeks help with two problems involving charged balls. The first problem involves a fixed conducting ball with charge q1=3x10^-6C attracting a second ball with charge q2, resulting in a force of 13.5N, and later repelling with a force of 0.9N after being connected by a wire. The charge of the second ball is determined to be -5x10^-6C, and the separation distance is 0.10m. The second problem asks for the position where a third particle experiences no net force between two charged particles. Participants clarify the equations involved and the steps to isolate q2, emphasizing the importance of understanding the forces and charges involved. The conversation focuses on solving these electrostatic problems accurately.
thebe39
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Hi, I have an exam coming up soon. But I thought I would clear my doubts ahead of time.

1. A fixed conducting ball has charge q1= 3X10^-6C. An identical ball with charge q2 is held at a distance x away from q1. The two balls attract each other with a force of 13.5N. The balls are then connected by a conducting wire. After the wire is removed, the balls repel each other with a force of 0.9N.

a) what was the charge of q2 of the second ball?
b) What is the separation x between the balls?


2. At what position between particles 1 and 2 will particle 3 experience no net force?
q1= 2x10-9 q2=3x10-9 q3= -2x10-9. The distance between particle 1 and 2 is 0.04.


Thanks
 
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Welcome to PF.

Where did you get stuck on the first problem?
 


well I know that in the beginning the q2 should have a - charge since it is attracted to the q1 and then after the wire it should become -. I'm not really sure how to isolate for
q2 from the equation.
 


thebe39 said:
well I know that in the beginning the q2 should have a - charge since it is attracted to the q1 and then after the wire it should become -. I'm not really sure how to isolate for
q2 from the equation.

If the charge from both is the same it will be repulsive.

That means in the first you a force made up of the given charge times the unknown. The second force is derived from the average of the 2 charges squared.

F = kq1*q2/r2

After

F = k(q1 + q2)2/(4*r2)

Where (q1 + q2) / 2 is the average that gets squared.
 


how did you get F=Kq(q1+q2)/(4*r^2) ? where does the extra q come from?
 


btw the answer to this problem is apparently
a) -5x10-6 C
b) 0.10m

But I really just want to understand what I'm doing with the steps.
 


thebe39 said:
how did you get F=Kq(q1+q2)/(4*r^2) ? where does the extra q come from?

Sorry, there is no extra q. I noticed the typo and edited it apparently after you picked up the post.
Note that the correct formula is

F = k(q1 + q2)2/(4*r2)
 


So I'm going to let r^2 equal each other

kq1q2/F = k(q1 +q2)^2/F(4)

So I isolate for q2 right?

or do I say that kq1q2/F = k(q1 +q2 divide by 2)^2/F(4)
 


thebe39 said:
So I'm going to let r^2 equal each other

kq1q2/F = k(q1 +q2)^2/F(4)

So I isolate for q2 right?

or do I say that kq1q2/F = k(q1 +q2 divide by 2)^2/F(4)

The factor of 4 came from q=q1+q2 divide 2, so your first equation is correct, but not your second. Note that the two "F"'s are different: in kq1q2/F, F=-13.5 N, but in k(q1+q2)^2/4F, F=0.9 N.

Otherwise, you're on the right track. Cancel out the k, expand the right side, and you'll get a quadratic equation. It should be easy from there. Remember that q1, F, and the other F are all given.
 
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