MHTCET 19 Apr 2026 Shift 2 Paper

Concept:Electrostatic pressure on the charged shell and projected area determine the net repulsive force between the two hemispheres.Explanation:A uniformly charged thin spherical shell experiences an outward electrostatic pressure.For surface charge density $\sigma$, the electrostatic pressure is $P = \frac{\sigma^2}{2\epsilon_0}$.Consider a small area $dA$ on one hemisphere. The outward force on it is $dF = P\,dA$.Only the component along the axis of the hemispheres contributes to separating them.If $\theta$ is the angle between the outward normal and this axis, the useful component is $dF_{\parallel} = P\,dA\cos\theta$.Integrating over the whole hemisphere gives $F_{\mathrm{rep}} = P \int dA\cos\theta$.The integral $\int dA\cos\theta$ is the projected area of the hemisphere on its circular base, equal to $\pi R^2$.Therefore, $F_{\mathrm{rep}} = \frac{\sigma^2}{2\epsilon_0} \pi R^2 = \frac{\pi\sigma^2 R^2}{2\epsilon_0}$.The pressing force $F$ must balance this repulsive force, so $F$ is proportional to $\frac{1}{\epsilon_0}\sigma^2 R^2$.Answer:$F \propto \frac{1}{\epsilon_0}\sigma^2 R^2$