Electrostatics: Two charged balls with attraction

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Homework Help Overview

The discussion revolves around electrostatics, specifically involving two charged balls that exhibit attraction and repulsion based on their charges and distance. The original poster presents two problems: determining the charge of a second ball and finding the position where a third particle experiences no net force between two other charged particles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the nature of the charges involved, with some noting that the second ball must have a negative charge due to the attraction to the first ball. There are attempts to isolate the charge of the second ball using equations related to electrostatic force.

Discussion Status

Participants are actively engaging with the problem, questioning the derivation of equations and discussing the implications of charge interactions. Some guidance has been provided regarding the correct formulas and the relationships between the forces involved, but no consensus on the final answers has been reached.

Contextual Notes

There is mention of specific values for charges and forces, as well as a distance between particles, which may influence the calculations. The participants are navigating through the complexities of the equations without a complete resolution to the problems presented.

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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