Force Between Two Hemispherical Shells of Uniformly Charged Spherical Shell

To determine the force $F$ required to hold the two hemispherical shells together, we calculate the net electrostatic repulsive force exerted by the charged shell on one of its hemispheres.

The electrostatic pressure $P$ acting outward on the surface of a uniformly charged thin spherical shell with surface charge density $\sigma$ is given by:
$P = \frac{\sigma^2}{2\varepsilon_0}$

Consider one of the hemispherical shells with its symmetry axis oriented along the $x$-axis. The element of surface area $dA$ on the spherical surface at an angle $\theta$ with respect to the $x$-axis is:
$dA = 2\pi R^2 \sin\theta \, d\theta$

The component of electrostatic force acting on this area element along the $x$-axis (the axis along which the force $F$ is applied) is:
$dF_x = P \cos\theta \, dA = \left(\frac{\sigma^2}{2\varepsilon_0}\right) \cos\theta (2\pi R^2 \sin\theta \, d\theta)$

Integrating $dF_x$ over the entire hemisphere ($\theta$ going from $0$ to $\frac{\pi}{2}$):
$F_{net} = \frac{\pi \sigma^2 R^2}{\varepsilon_0} \int_{0}^{\pi/2} \sin\theta \cos\theta \, d\theta$

Evaluating the integral:
$\int_{0}^{\pi/2} \sin\theta \cos\theta \, d\theta = \left[ \frac{\sin^2\theta}{2} \right]_{0}^{\pi/2} = \frac{1}{2}$

Substituting this back into the expression for $F_{net}$:
$F_{net} = \frac{\pi \sigma^2 R^2}{2\varepsilon_0}$

Alternatively, using the concept of projected area, the total force due to uniform pressure $P$ along the axis of symmetry is equal to the pressure multiplied by the projected area of the hemisphere ($\pi R^2$):
$F_{net} = P \times (\pi R^2) = \frac{\sigma^2}{2\varepsilon_0} \cdot \pi R^2 = \frac{\pi \sigma^2 R^2}{2\varepsilon_0}$

To keep the two hemispherical shells together, the applied force $F$ must balance this electrostatic repulsive force:
$F = F_{net} = \frac{\pi}{2\varepsilon_0} \sigma^2 R^2$

Thus, the force $F$ is proportional to:
$F \propto \frac{1}{\varepsilon_0} \sigma^2 R^2$

Correct Option: A