Two unequal charges repel, suspended from same point.

In summary, two small, identical spheres of mass m each are attached to thin light threads of length L each and hung from the same point. When charges q1 and q2 are placed on upon the spheres(with both charges having the same sign), the two spheres repel each other and, upon reaching equilibrium, hang at angles \theta and \theta' from the vertical. If q1= 3q2, then \theta' is equal to \theta. This is because the electrostatic forces caused by the charges are in line with the distance between the spheres, and assuming the charges are at the same level, the forces are horizontal. By equating the forces caused by the charges and the x-component of the two tension forces
  • #1
1. Two small, identical spheres of mass m each are attached to thin light threads of length L each and hung from the same point. When charges q1 and q2 are placed on upon the spheres(with both charges having the same sign), the two spheres repel each other and, upon reaching equilibrium, hang at angles [tex]\theta[/tex] and [tex]\theta[/tex]' from the vertical.
a) If q1= 3q2, what is [tex]\theta[/tex]' in terms of [tex]\theta[/tex]? (Hint: Consider Newton's Third Law)
b) Draw a free-body diagram for either sphere, and hence find the distance x between the spheres in terms of q1, q2, L, m, and any necessary physical constants. Assume that [tex]\theta[/tex] is small such that [tex]\theta[/tex] [tex]\approx[/tex] Sin[tex]\theta[/tex] [tex]\approx[/tex]Tan[tex]\theta[/tex]




Homework Equations


F= kq1q2/d2


The Attempt at a Solution


In my first attempt at the problem, I assumed the forces caused by the charges were perpendicular to the vertical and thus got the answer Sin[tex]\theta[/tex]/Sin[tex]\theta[/tex]' = 1. I think this answer is incorrect, but I got it by equating the two forces caused by the charges, and hence equating the x-component of the two tension forces.

I guess the correct way to do it would be to take the electrostatic forces as being in line with the distance, but I can't seem to figure out the angle between the force and the horizontal. I am also unsure as to whether I should use the statement given in part b as how small [tex]\theta[/tex] is. Any help would be greatly appreciated.
 
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  • #2
Welcome to PF!

Hi Darkwolf312! Welcome to PF! :smile:

(hav a theta: θ :wink:)
Darkwolf312 said:
Assume that [tex]\theta[/tex] is small such that [tex]\theta[/tex] [tex]\approx[/tex] Sin[tex]\theta[/tex] [tex]\approx[/tex]Tan[tex]\theta[/tex]

In my first attempt at the problem, I assumed the forces caused by the charges were perpendicular to the vertical and thus got the answer Sin[tex]\theta[/tex]/Sin[tex]\theta[/tex]' = 1. I think this answer is incorrect, but I got it by equating the two forces caused by the charges, and hence equating the x-component of the two tension forces.

I guess the correct way to do it would be to take the electrostatic forces as being in line with the distance, but I can't seem to figure out the angle between the force and the horizontal.

Yes, you're correct, the force is in line with the distance :smile:

but you should assume that the charges are at the same level, so that the forces are horizontal …

that's what the question is trying to get at :rolleyes: when it talks about θ being small! :wink:
 

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