An uniformly charged thin spherical shell of radius 'R' has uniform surface charge density '\(σ\)'. It is made of two hemispherical identical shells held together by pressing them with force 'F' as shown. F is proportional to [\(ε_0\) = permittivity of free space]
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The outward electrostatic pressure on a charged surface is sigma squared over 2 epsilon 0; multiply by the cross-section area of a hemisphere.
Step 1: Field on the surface
The field just outside the shell is $E = \dfrac{\sigma}{\varepsilon_0}$ and inside it is zero, so the average field acting on the surface charge is $\dfrac{\sigma}{2\varepsilon_0}$.
Step 2: Force on a small element
A small element of charge $\sigma\,dA$ feels a force $\sigma\,dA\cdot\dfrac{\sigma}{2\varepsilon_0}$, directed radially outward.
Step 3: Add the components
Adding the components along the axis of the hemisphere gives the same result as the projected area $\pi R^2$: $F = \dfrac{\sigma^2}{2\varepsilon_0}\pi R^2$.
Step 4: Result
$F = \dfrac{\pi\sigma^2R^2}{2\varepsilon_0}$, which is proportional to $\dfrac{1}{\varepsilon_0}\sigma^2R^2$, option (C).
Final Answer:
F is proportional to sigma^2 R^2 / epsilon_0. This is option (C).
\[ \boxed{\text{(C) }\frac{1}{\varepsilon_0}\sigma^2R^2} \]
