Electrostatic Equillibrium Question

  • Thread starter Thread starter vg19
  • Start date Start date
  • Tags Tags
    Electrostatic
AI Thread Summary
The discussion revolves around calculating the common charge on three identical Styrofoam balls suspended in equilibrium, forming an equilateral triangle. The user initially assumes the angles between the strings and the vertical are 60 degrees but is advised that this is incorrect. Geometry is emphasized as crucial for determining the correct angles, with suggestions to use the properties of isosceles triangles and the inscribed circle of the equilateral triangle. The user is encouraged to find the radius of the inscribed circle to accurately calculate the angle the strings make with the vertical. The conversation highlights the importance of correct geometric assumptions in solving electrostatic equilibrium problems.
vg19
Messages
67
Reaction score
0
Hey,

Here is my question:
Three identical small Styrofoam balls (m = 2.05 g) are suspended from a fixed point by three nonconducting threads, each with a length of 45.5 cm and with negligible mass. At equilibrium the three balls form an equilateral triangle with sides of 28.4 cm. What is the common charge q carried by each ball?

This is what I have done.

I made my diagram 3d and did a FBD for 1 ball (Tension up and at an angle, mg going down, and Fe going to the left) As a result, solving for the Forces in the x-direction I had
Ftcos60 = Fe
and solving in the y direction
Ftsin60 = mg

I divided EFy by EFx to get rid of the tension force and then solved for Fe. I then put that into the coulomb law equation and made both q charges the same so it was q^2 and solved for q. I am not getting the right answer though. I think my assumption of the angles being 60 may be wrong. Can anybody help out?

Thanks
 
Physics news on Phys.org
The angle that the strings make with the vertical is not 60 degrees (nor is it 30 degrees!).

You're going to have to do some geometry to figure out the angle but it is simplified by noting the triangle with vertices at each of the charges is equilateral and the triangles with the vertices at the support and pairs of points is isosceles. It may also help to note that the center of the equilateral triangles is located 1/3 of the way from the base to the opposite vertex.
 
I tried using the cosine law, and for the isosoleces triangle I am getting a top angle of 36.4degrees and the two bottom ones as 71.8degrees, but I am still not getting the right answer...Am I still doing something wrong?
 
What you are looking for is not the angle between a pair of strings but the angle between the vertical and a string.
 
So would I just divide the 36.4 by 2?
 
The easiest way to do the geometry is to first find the radius of a circle in which the equilateral triangle can be inscribed. Then you can find the angle the string makes with the vertical using the ratio of that radius to the length of the string.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top