Coulomb's Law, what charge will make the two charges in static eq.

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Homework Help Overview

The problem involves determining the charge q1 that will allow charge q2 to be in static equilibrium with a given charge Q of -3.7 nC. The context is rooted in Coulomb's Law and the forces acting between charged particles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the distances of the charges and the forces acting on them. There is confusion regarding the application of the inverse square law and the implications of squaring the charge in calculations.

Discussion Status

The discussion is active with participants sharing their reasoning and questioning assumptions about the relationships between the charges and forces. Some have offered insights into the effects of distance on force and the necessary adjustments to charge values, while others express uncertainty about their calculations.

Contextual Notes

There appears to be some misunderstanding regarding the units of force and the application of the inverse square law in the context of this problem. Participants are also navigating the implications of distance on the forces involved.

rocapp
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Homework Statement


In the figure below the charge in the middle is
Q = -3.7 nC. For what charge
q1 will charge q2
be in static equilibrium?

Homework Equations


F = (K*q1*q2)/(r^2)

The Attempt at a Solution



I'm not sure where to go after acknowledging that Fq1->q = -Fq2->q
 

Attachments

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rocapp said:

Homework Statement


In the figure below the charge in the middle is
Q = -3.7 nC. For what charge
q1 will charge q2
be in static equilibrium?

Homework Equations


F = (K*q1*q2)/(r^2)

The Attempt at a Solution



I'm not sure where to go after acknowledging that Fq1->q = -Fq2->q
attachment.php?attachmentid=56732&d=1363310960.png



Let's see ...

q1 is twice the distance from q2 that Q is from q2, and we have an inverse square law.
 
So that means the force is (3.7^2) = ~14 nC?
 
rocapp said:
So that means the force is (3.7^2) = ~14 nC?
The unit for force is not Coulombs .

Why would you square the charge anyway ?
 
I misunderstood the inverse square law.

F=q1=q2=k*q1*q2/r^2 = k*q1*Q/(r/2)^2

(9x10^9)*(q1*q2)/400 = (9x10^9)*q1*(3.7)/100

q1*q2/400 = q1*0.037
q2 = 14.8 N = F

Is that correct?
 
rocapp said:
I misunderstood the inverse square law.

F=q1=q2=k*q1*q2/r^2 = k*q1*Q/(r/2)^2

(9x10^9)*(q1*q2)/400 = (9x10^9)*q1*(3.7)/100

q1*q2/400 = q1*0.037
q2 = 14.8 N = F

Is that correct?
Doubling distance reduces force to 1/4 , so to compensate for that, charge must be 4 times what is would be at Q . (Of course with opposite sign.)

So, is 14.8 = (4)(3.7) ?

... Yes. So you're O.K.
 
Oh! Wow, that's a lot simpler than I was trying to make it. Thanks!
 

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