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]
Show Hint
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: Understand the figure
The figure shows a uniformly charged spherical shell split along a plane into two identical hemispheres, held together by equal and opposite forces \(F\) pressing them toward each other.
Step 2: Electrostatic pressure
Each small patch of the surface is repelled by the rest of the charge. The outward force per unit area (electrostatic pressure) on a surface charge \(\sigma\) is \(\dfrac{\sigma^2}{2\varepsilon_0}\).
Step 3: Net force on a hemisphere
The outward forces on the hemisphere add up to the pressure times the area of the flat circular cross-section, \(\pi R^2\):
\[ F = \frac{\sigma^2}{2\varepsilon_0}\times\pi R^2 \]
Step 4: Result
So \(F \propto \dfrac{1}{\varepsilon_0}\sigma^2R^2\), option (C). The pressing force must balance this repulsion. Options with \(R^2\) in the denominator, or with a single power of \(R\), do not follow from pressure times area.
Final Answer:
F is proportional to sigma^2 R^2 / epsilon_0. This is option (C).
\[ \boxed{\text{(C) }\frac{1}{\varepsilon_0}\sigma^2R^2} \]
