Electrical force between two objects

[IAmCurious]

I tried to solve the following problem:
There are 2 objects . The object m1 with charge q and the object m2 with charge q.(same charge).The object m2 is connected with a rope to the ceiling.

at the initial state , m1 and m2 were on the same height.
When the object m2 is placed in distance d from the object m1, the following situation occurs:

[url]http://i.stack.imgur.com/KhKjc.png[/url]

Please ignore the force of gravity between the masses and ignore any kind of friction.
just consider the tension , electrical force and mg.

You have to find x.

You can use the parameters k(Coulomb's constant), q , L , d , mg.

I was able to find only 2 equations that would help to solve the problem but one is still missing.

 

[willem2]

I tried to solve the following problem:
There are 2 objects . The object m1 with charge q and the object m2 with charge q.(same charge).The object m2 is connected with a rope to the ceiling.

at the initial state , m1 and m2 were on the same height.
When the object m2 is placed in distance d from the object m1, the following situation occurs:

[url]http://i.stack.imgur.com/KhKjc.png[/url]

Please ignore the force of gravity between the masses and ignore any kind of friction.
just consider the tension , electrical force and mg.

You have to find x.

You can use the parameters k(Coulomb's constant), q , L , d , mg.

I was able to find only 2 equations that would help to solve the problem but one is still missing.

2 equations should be enough. The unknowns are x and the tension in the rope.

 

[IAmCurious]

I checked my equations and they are not correct. I wasn't able to calculate the components of the electrical force.
Could you please write the equations you have found ?

The equations that I found are :

1) r ^ 2 = (L-Lcos a) ^ 2 + (x+d) ^ 2
(r is the distance between the two objects)

2) sin a = x / L

3) as you can see ΣFx = 0 and ΣFy = 0 (The object m2),
but without the components of the electrical force I wansen't able to calculate anything.