A thin insulating thread of length L is attached to a point on the perimeter of a thin disk.

Disk is has radius R< L and is charged with a uniform surface density, σ>0. At the end of
the thread there is a small ball mass m and charge q>0. The mass happens to be located
on disk's axis. What is the mass of the ball?

Figure the Electric field from the charged disk along the axis at a distance z, given R

http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/elelin.html#c3

Now, given that expression, then you know from the geometry with the angle theta the string makes from disk,
tanTheta=Eq/mg
But CosineTheta=R/L
and sinTheta=sqrt((L^2-R^2)/L^2) or
sinTheta=sqrt(1-(R/L)^2)
but sinTheta/cosTheta= tan Theta, so

Eq/mg=sqrt(1-(R/L)^2) *L/R
Eq/mg=sqrt((L/R)^2-1)
and now you can solve for L

Remember in your E expression, z=Sqrt(L^2-R^2)

As you can tell, some algebra will be required. Get a nice pad of paper.

To find the mass of the ball, we need to consider the forces acting on it and use Newton's second law of motion.

First, let's analyze the forces acting on the ball:

1. Gravitational force (Fg): This force is directed downward and is given by Fg = mg, where m is the mass of the ball and g is the acceleration due to gravity.

2. Electrostatic force (Fe): Since the ball has a charge q, it will experience an electrostatic force due to the charged disk. The electrostatic force is given by Fe = k * q * Q / r^2, where k is Coulomb's constant (k = 9 x 10^9 N * m^2 / C^2), Q is the total charge on the disk, and r is the distance between the ball and the center of the disk.

Now, let's consider the equilibrium condition of the system. Since the ball is in equilibrium, the net force acting on it must be zero. Therefore, we can write:

Fg + Fe = 0

Substituting the expressions for Fg and Fe, we get:

mg + k * q * Q / r^2 = 0

Now, rearranging the equation, we can solve for the mass of the ball:

m = (-k * q * Q) / (g * r^2)

So, the mass of the ball is given by m = (-k * q * Q) / (g * r^2).