JEE Advanced · Physics · 17. Electrostatics
A uniformly charged thin spherical shell of radius \(R\) carries uniform surface charge density of \(\sigma\) per unit area. It is made of two hemispherical shells, held together by pressing them with force \(F\) (see figure). \(F\) is proportional to

- A \(\frac{1}{\varepsilon_0} \sigma^2 R^2\)
- B \(\frac{1}{\varepsilon_0} \sigma^2 R\)
- C \(\frac{1}{\varepsilon_0} \frac{\sigma^2}{R}\)
- D \(\frac{1}{\varepsilon_0} \frac{\sigma^2}{R^2}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{\varepsilon_0} \sigma^2 R^2\)
Step-by-step Solution
Detailed explanation
Electrical force per unit area \(=\frac{1}{2} \varepsilon_0 E^2\)
\(
=\frac{1}{2} \varepsilon_0\left(\frac{\sigma}{\varepsilon_0}\right)^2=\frac{\sigma^2}{2 \varepsilon_0}
\)
Projected area \(=\pi R^2\)
\(\therefore\) net electrical force \(=\left(\frac{\sigma^2}{2 \varepsilon_0}\right)\left(\pi R^2\right)\)
In equilibrium, this force should be equal to the applied force.
\(\therefore \quad F=\frac{\pi \sigma^2 R^2}{2 \varepsilon_0}\)
or \(\quad F \propto \frac{\sigma^2 R^2}{\varepsilon_0}\)
\(\therefore\) The correct option is (a).
\(
=\frac{1}{2} \varepsilon_0\left(\frac{\sigma}{\varepsilon_0}\right)^2=\frac{\sigma^2}{2 \varepsilon_0}
\)
Projected area \(=\pi R^2\)
\(\therefore\) net electrical force \(=\left(\frac{\sigma^2}{2 \varepsilon_0}\right)\left(\pi R^2\right)\)
In equilibrium, this force should be equal to the applied force.
\(\therefore \quad F=\frac{\pi \sigma^2 R^2}{2 \varepsilon_0}\)
or \(\quad F \propto \frac{\sigma^2 R^2}{\varepsilon_0}\)
\(\therefore\) The correct option is (a).
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