Three charges of mass $m$ and charge $q$ are tied to massless ropes of length $l_!$, and hung from a single, fixed point. At equilibrium, the charges form an equilateral triangle in a horizontal plane with sides of length $l_2$
Attempted solution
It seems to me that it is sufficient to consider a single charge.
The component of tension (call it $T$) in the horizontal plane will bisect the $60^{\circ}$ angle (formed by the "axis" on which lie the components of force due to the other two charges on this charge in the horizontal plane). For convenience, we will let the direction of this component define the $x$-axis.
The $x$-axis component of the magnitude of the force due to the other two charges on this charge is given by Coulombs law, I get:
$$2\bigg[ k_e\frac{||q||^2}{(l_2)^2}\cos(30^{\circ})\bigg]$$
Call the angle formed by the massless ropes of length $l_1$ and the horizontal plane by $\theta$ (this angle is "orthogonal" to the plane). We can then write:
$$T\cos(\theta) = 2\bigg[ k_e\frac{||q||^2}{(l_2)^2}\cos(30^{\circ})\bigg]$$
Re-arrange the expression to solve for $q$:
$$\sqrt{\frac{(l_2)^2T\cos(\theta)}{2k_e\cos(30^{\circ})}} = ||q||$$
I'm having trouble moving forward from here. I know that $T = mg + F_c$ (where $F_c$ is the force due to the other two charges), but then I'm stuck in a loop of attempting to solve for $q$.
I solved this before, and got $110nC$, which I understand to be correct. Unfortunately, I lost my solution and have failed to reconstruct it.
