# Finding the Magnitude of charge from a Gold-leaf electroscope

I think the question should be, is it even possible to find the charge of a body placed on the Gold-leaf electroscope. There is definitely a quantitative relation between the angle of separation and the quantity of charge

I thought for a while what happens when a charged object is placed on a Gold-leaf electroscope. It is clear that the charge is equally distributed among the leaves as well as the whatever-conducting-material-is-there-attached-to-the-leaves.

Assuming that the charges are equally distributed among the three, the next step should be finding the Coloumbic forces between them, knowing that this is the force that probably causes a torque which is what results in separation angle.

But wait, If there are Coloumbic forces and these forces work over a large range and are very strong then why isn't the leaves separating such that the repulsion is minimum(subquestion 1). What I got in my mind is the Valence shell electron pair repulsion theory, or VSEPR theory where you get the orientation of the electrons such that the repulsion is minimum, then why isn't the angle $$180$$?

A possible culprit is the conducting part of the electroscope which may also have electrons and thus repel both the gold leaves. But assuming that all three have the same no: of electrons then shouldn't the angle always be $$120$$. It definitely may not be the effect of gravity as the force of gravity is weak compared to the electrostatic forces

Then there's the elephant in the room, how do you integrate this knowing every point on the is effected by every other point on the leaves

TIA

To see this imagine the charged object was the same size and material as the conducting part of the electroscope and has charge $$2Q$$. Then half the charge would go onto the electroscope and half would stay on the charged object.
If we used an object of charge $$3Q$$ made of two copies of the electroscope - then when it was placed on the original electroscope, $$Q$$ would go onto it and $$2Q$$ would remain on the charged object.
In both the above situations $$Q$$ has gone onto the original electroscope, so the readings would be the same, but the charges of the two objects were different.
As $$T \sin{\theta} = F_e = k Q^2 / x^2$$ and $$T \cos{\theta} = m g$$, we get $$Q = x \sqrt{m g tan{\theta} / k}$$