Solution

If we solve this problem using the equilibrium condition

$$F_k=mg,$$

you will get an incorrect answer, because such an equilibrium would be unstable.

To verify stability, it is necessary to displace the ball by a small angle $\alpha\longrightarrow0$; $(\sin{\alpha}\approx\alpha)$ and consider for which $q$ a restoring force will arise.

It should be noted that for small displacements, the Coulomb force can be considered constant

$$F_k=\text{const}=\frac{1}{4\pi\varepsilon_0}\frac{qQ}{4R^2}$$

Let us consider the projections of forces onto the $Ox$ axis, parallel to the tangent to the sphere at the point of contact with the ball (inclined at an angle $\alpha$ to the horizontal):

$$(mg)_x=-mg \cdot \sin{\alpha}$$

$$(N)_x=0$$

$$(F_k)_x=F_k \cdot \sin{\alpha /2}$$

In this case, the Coulomb force must be significantly greater than $mg$, i.e., the ball will not fall down but can roll down the wall, which means the equilibrium will be stable if the sum of the projections of all forces onto $Ox$ is greater than zero

$$F_k \cdot \sin{\alpha /2}-mg \cdot \sin{\alpha}=0$$

$$F_k=mg \cdot \frac{\alpha}{\alpha /2}$$

$$\frac{1}{4\pi\varepsilon_0}\frac{qQ}{4R^2}=2mg$$

$$q=\frac{32\pi\varepsilon_0mgR^2}{Q}$$