Two Charged Masses Suspended From Strings

  • Thread starter Thread starter Geromy
  • Start date Start date
  • Tags Tags
    Charged Strings
AI Thread Summary
Two charged masses suspended from strings experience forces due to electric repulsion and gravity, forming a right triangle with the tension in the string. The equilibrium condition leads to the relationship between the distance x between the charges and the other parameters involved. By applying Coulomb's Law and trigonometric identities, the equation for x is derived as x = ((q^2L)/(2∏ε0mg))^(1/3). The discussion highlights the importance of understanding the balance of forces and the geometry of the system. This problem illustrates the interplay between electrostatic and gravitational forces in a suspended system.
Geromy
Messages
8
Reaction score
0

Homework Statement



Two similar tiny balls of mass m are hung from silk threads of length L and carry equal charges q. The angle formed by the two strings is bisected by an imaginary line, forming angle θ. Assume that θ is so small that tan θ can be replaced by its approximate equal, sin θ. The distance between the two charged masses, at equilibrium, is x. Show that, for equilibrium,

x = ((q2L)/(2∏ε0mg))1/3


Homework Equations



Clearly, Coulomb's Law is a vital piece of this problem, and other sources that I've checked suggest such things as the constancy of the ratio of the sides of a triangle divided by the sine of that side's opposite angle and combining the electric force and gravitational force into the components of the force of tension.



The Attempt at a Solution



Despite having a couple of ideas about how to start, I haven't made much progress. I noticed that there is an identical problem here (https://www.physicsforums.com/showthread.php?t=305517) that suggested the constant sine ratio rule, but I can't for the life of me get to the right answer. Any help would be appreciated.
 
Physics news on Phys.org
Haha, so, I figured it out! I thought I would post here in case anyone wants an explanation.

Consider the forces acting on a single one of the charged masses: the force due to the charges is kq2/x2 - x is the radius between the two charges, since it's the distance they're separated by.

The force due to gravity is also pretty straightforward: it's simply mg.

Since these two forces are perpendicular, they form two sides of a right triangle. The hypotenuse of this triangle is equal to the final force acting on the charged mass, the force of tension. This triangle contains angle θ, since the force of tension is in the direction of the silk thread.

Since now we've reintroduced θ, we can start using trigonometry. Tan θ = Opp/Adj, which in this case is our force due to electric charge (F) over the force due to gravity (W), or F/W.

However, Tan θ = Sin θ, which equals x/2L

when you combine equations, you get 2kq2L = mgx3, and from here it's a simple matter of algebra to get to the final equation.
 
Well done!

Welcome to PF!

ehild
 
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