An uniformly charged thin spherical shell of radius 'R' has uniform surface charge density '$\sigma$'. It is made of two hemispherical identical shells held together by pressing them with force 'F' as shown. F is proportional to
[$\epsilon_0$ = permittivity of free space]

The electrostatic pressure $P$ on the surface of a uniformly charged thin spherical shell is given by:

$P = \dfrac{\sigma^2}{2\epsilon_0}$

Consider a small area element $dA$ on the surface of one of the hemispheres. The outward electrostatic force on this element is:

$dF = P dA = \dfrac{\sigma^2}{2\epsilon_0} dA$

By symmetry, the net force on the hemisphere will be directed along the axis perpendicular to its circular base. The component of the force along this axis is:

$dF_{\perp} = dF \cos\theta = P dA \cos\theta$

where $\theta$ is the angle between the radially outward normal and the axis of symmetry.

Integrating this over the entire surface of the hemisphere gives the total repulsive force:

$F_{rep} = \int P dA \cos\theta = P \int dA \cos\theta$

The term $\int dA \cos\theta$ represents the projected area of the hemisphere onto the plane of its base, which is simply the area of a circle of radius $R$:

$\int dA \cos\theta = \pi R^2$

Substituting this back, we get the net repulsive force:

$F_{rep} = \left(\dfrac{\sigma^2}{2\epsilon_0}\right) (\pi R^2) = \dfrac{\pi \sigma^2 R^2}{2\epsilon_0}$

To hold the two hemispheres together, the external applied force $F$ must balance this electrostatic repulsion:

$F = \dfrac{\pi \sigma^2 R^2}{2\epsilon_0}$

From this expression, it is clear that the force $F$ is proportional to $\dfrac{\sigma^2 R^2}{\epsilon_0}$.

Answer: $\dfrac{1}{\epsilon_0}\,\sigma^2 R^2$