At what point is the system in equilibrium?

AI Thread Summary
The discussion centers on determining the equilibrium point for a proton placed between two fixed charges: +8q at the origin and -2q at x=L. The correct equilibrium point is located to the right of the -2q charge, specifically L units away from it, which translates to 2L from the origin. A proposed equation setup using the distances from the charges led to confusion, as it measured x from the -2q charge instead of the origin. Ultimately, both methods yield the same physical location for equilibrium, but require careful interpretation of the variables used. Understanding the reference points for distance is crucial for consistent results in such problems.
catch22
Messages
62
Reaction score
0

Homework Statement


This is more of a math question but:

There are two particles fixed in place: a particle of charge +8q at the origin and a particle of charge -2q
at x=L. At what point (other than infinitely far away) can a proton be placed so that it is in equilibrium

Homework Equations

The Attempt at a Solution


The point should be at the right of the -2q charge, so it is farther away from the charge +8q.

upload_2015-10-26_7-33-56.png


upload_2015-10-26_7-34-40.png


upload_2015-10-26_7-35-2.png


diagram:
upload_2015-10-26_7-35-33.png


that was the answer, but would it be possible to use

8q / (L+x)2 = 2q / (x)2

where L is the distance between the charges 8q and -2q.

I get x = L using this set up instead.

what went wrong? Shouldn't I get the same answer?
 
Physics news on Phys.org
catch22 said:
that was the answer, but would it be possible to use

8q / (L+x)2 = 2q / (x)2

where L is the distance between the charges 8q and -2q.

I get x = L using this set up instead.

what went wrong? Shouldn't I get the same answer?
The answers are the same! Realize that in the second approach you are measuring x from the second charge, not the origin.
 
Doc Al said:
The answers are the same! Realize that in the second approach you are measuring x from the second charge, not the origin.
Ah, so the point is L away from the -2q and 2L away from the origin.
 
catch22 said:
Ah, so the point is L away from the -2q and 2L away from the origin.
Exactly. What's confusing you is that in the second method, "x" stands for "unknown", not the x-coordinate of the final location. You'll have to translate to that to report your answer, if you want them to match.
 
  • Like
Likes catch22
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top