Balancing Charges on Identical Spheres

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Homework Help Overview

The problem involves two identical metal spheres with initial charges that experience both attractive and repulsive forces when placed a certain distance apart. The context is centered around understanding the behavior of electric charges and the application of Coulomb's Law.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculations related to the forces experienced by the spheres and the resulting charges after they come into contact. There is confusion regarding the correct expressions for the charges after they balance and how these relate to the forces observed.

Discussion Status

Some participants are exploring the implications of the charge distribution after the spheres come into contact, while others are questioning the assumptions made about the nature of the charges and the resulting forces. There is no explicit consensus on the correct approach to the problem.

Contextual Notes

Participants are grappling with the definitions of the charges on the spheres after they have been brought together and are uncertain about the total charge conservation. The discussion reflects a mix of interpretations regarding the relationships between the charges and the forces involved.

Tridius
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Homework Statement




Two identical small metal spheres initially carry charges q_1 and q_2. When they're x=1.0m apart, they experience a 2.5N attractive force. Then they're brought together so charge moves from one to the other until they have the same net charge. They're again placed x=1.0m apart, and now they repel with a 2.5N force.

Homework Equations



F=kq1q2/r^2

The Attempt at a Solution



Alright, so this question seemed pretty easy but my answer isn't making any sense. First I solved for the net charge on the spheres after they had discharged,

F=kq3^2/r^2 ... q3=sqrt(Fr^2/k) = 1.667*10^-5C

Next I assumed that since the two spheres had balanced their charges, the original charges were 1.667*10^-5 +/- x.

Putting this back into the Coulomb's Law equation I got

F=k(1.667*10^-5 - x)(1.667*10^-5 + x)/r^2

But when I solved for x and then tried to sub it back into q1=(1.667*10^-5 - x) and q2=(1.667*10^-5 + x), both q1 and q2 were positive, which would not result in an attractive force.

Help please. =)
 
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Hi Tridius. welcome to PF.
In the first case, F = k*q1*q2/d^2.
In the second case, charge on each sphere is (q1 - q2)/2
So F= k*(q1 - q2)^2/d^2
Can you proceed now?
 
Thanks a lot, this should really help. What I don't understand though is that if the charges are now (q1-q2)/2, then why is the force k(q1-q2)^2/d^2? Where did the 2 in the denominator go?
 
Tridius said:
Thanks a lot, this should really help. What I don't understand though is that if the charges are now (q1-q2)/2, then why is the force k(q1-q2)^2/d^2? Where did the 2 in the denominator go?
The force should be
F = k(q1-q2)^2/4*d^2
 
I don't understand why each charge is (q1-q2)/2.
If it was, then the total charge would be 2*(q1-q2)/2 = q1-q2.
But we know the total charge is q1 + q2.
Looks like the charge on each must be (q1 + q2)/2.
 
Delphi51 said:
I don't understand why each charge is (q1-q2)/2.
If it was, then the total charge would be 2*(q1-q2)/2 = q1-q2.
But we know the total charge is q1 + q2.
Looks like the charge on each must be (q1 + q2)/2.
Since the force is attractive, the charges must be of opposite nature.
 

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