Find Coulomb Charge on Two Repelling 1g Masses | Homework Help

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The discussion revolves around calculating the charge on two equally charged 1g masses that repel each other, with one mass fixed. Participants clarify the use of gravitational and Coulomb's law equations, specifically addressing the variables involved, particularly the distance "r." It is established that "r" in the gravitational equation represents the distance from the center of the Earth to the object, while in Coulomb's law, it is the separation between the two charges. The conversation highlights the challenge of solving for charge and distance simultaneously due to having one equation with two unknowns. Ultimately, the need for additional information or clarification from the problem source is emphasized.
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Homework Statement


Two equally charged 1g masses repel each other. The lower mass is held fixed.

a)What is the charge on each mass for the Coulomb force to balance the gravitational force of the Earth on the upper mass?

b)How many elecetrons does this rep?

Homework Equations


F=Gm1m2/r2, where G=6.67x10-11
F=Kq1q2/r2, where K=9x109


The Attempt at a Solution


I know to find the charge on each mass we first use the gravitational equation and then use Coulombs law equation to find out the charge on each mass:

F=Gm1m2/r2
=(6.67x10-11)(1)(1)/r2

I'm pretty sure this is right but I'm just confused what r is. Could it be 5.0x10-11 which is the distance between an electron and proton?

b)This is simple. When you find out the coulomb charge, you use 1Coulomb=6.25x1018 electron charge.
 
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physics(L)10 said:

Homework Equations


F=Gm1m2/r2, where G=6.67x10-11
F=Kq1q2/r2, where K=9x109


I'm pretty sure this is right but I'm just confused what r is. Could it be 5.0x10-11 which is the distance between an electron and proton?
Look at your relevant equations. What distance does "r" represent in the first equation? What about the second equation?
 
I believe r in the first equation is between the centr of the Earth and the object, so would it just be as if the object is on the surface of the earth. And the 2nd equation is the distance between the two objects.
 
Correct. Note that, near the surface of the Earth, the gravitational force can be written as F = mg. In other words, GME/RE2 is replaced with g. Does this help?
 
Alright so let me attempt this:

F=mg=(1)(9.8)=9.8. This can then be used:

F=Kq1q2/r2

9.8=(9x109)(q1q2)/r2, but then I'm still confused with what r is. Is it what I said?
 
Actually, 1 g = 0.001 kg and that's the mass you should use. We have now sorted out the r's. Here r represents the separation between the two charges. Now since q1=q2, the equation that you wrote becomes

kq2/r2 = mg

This is one equation with two unknowns, q and r. To solve this for q (and r), we need either one more equation or one more value for the one of the unknowns. This is because different charges will give rise to different separations in the Earth's gravitational field. I don't see where one can get one more equation. Does the problem provide one more value?
 
Nope, I double-checked the problem and I wrote it exactly the same. Is it possible we can used substitution? Ex, solve for q and then plug it back into solve for r.
 
Solve what for q and then plug back in what? You only have one equation and two unknowns. If you do that to the one equation you have, you will end up with 0 = 0. Think of it this way. There is an infinity of pairs of q and r such that

mg = kq2/r.
 
Well I don't know what to do then lol.
 
  • #10
physics(L)10 said:
Well I don't know what to do then lol.
You could ask the person who asked you to solve this problem for a clarification.
 
  • #11
Alrighty, thanks a lot :)
 
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