Electric force between two point charges

[slugbunny]

Homework Statement

Two identical spheres with mass m are hung from silk thread with length L. Each sphere has the same charge, so q₁=q₂=q. The radius of each sphere is very small compared to the distance between the two spheres, so they may be treated as point charges. Show that if the angle θ is small, the equilibrium separation d between the spheres is

d=(q²L/2πε₀mg)^1/3

(Hint: If θ is small, tanθ=sinθ)

Homework Equations

F=ma=Eq
The masses are hung from thread connected at the same point so that it looks pendulum-esque and possible use of triangles.

The Attempt at a Solution

Well I started off with this equation
F=mg=Eq

then used symmetry
mg=2E_xq
mg=2E sin(θ) q
mg=2q/(4πε₀) (1/(1/2d)²) sin(θ) q

and sin(θ)=r/L so I got

mg=2q²/(4πε₀) (4/d²) (r/L)
mg=2q²/(πε₀d²) (d/2L)
mg=q²/(πε₀dL)

and d is...

d=q²/(πε₀Lmg)

I'm thinking that I would get the correct d if I lose the 2 from the electric field symmetry and switch the sin(θ) values (L/r instead of r/L), then I would get a d³ and an L in the numerator and it would work fine hahaha.

Thanks for help!

 

[Fightfish]

Well I started off with this equation
F=mg=Eq

then used symmetry
mg=2E_xq
mg=2E sin(θ) q
mg=2q/(4πε₀) (1/(1/2d)²) sin(θ) q

My first reaction would be "huh"? Who gave you the idea that mg = qE? In fact, the weight and electric force act in perpendicular directions. Clearly there must be another force that balances the system such that it is in equilibrium! That would be tension in the strings. So,
mg = T \,cos \theta
qE = T\, sin \theta
Alright, carry on.

 

[slugbunny]

Thanks, I totally forgot about tension U:

So I can solve for T and get

mg=qE cot(θ)

which gets me the right answer! Thanks so so much :)