# Electric force between two point charges

1. Jan 19, 2010

### slugbunny

1. The problem statement, all variables and given/known data

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)

2. Relevant 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.

3. The attempt at a solution

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

then used symmetry
mg=2Exq
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!

2. Jan 19, 2010

### Fightfish

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.

3. Jan 19, 2010

### 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 :)