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) $\frac{1}{\varepsilon_{0}} \sigma^{2} \mathrm{R}^{2}$

(B) $\frac{1}{\varepsilon_{0}} \sigma^{2} \mathrm{R}$

(C) $\frac{1}{\varepsilon_{0}} \frac{\sigma^{2}}{\mathrm{R}}$

(D) $\frac{1}{\varepsilon_{0}} \frac{\sigma^{2}}{\mathrm{R}^{2}}$

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Electrostatics

A uniformly charged thin spherical shell of radius

R

carries uniform surface charge density of

σ

per unit area. It is made of two hemispherical shells, held together by pressing them with force

F

(see figure).

F

is proportional to

Answer: (A)

The electric field just outside a uniformly charged spherical shell is:

E = \frac{σ}{ε _{0}}

The electrostatic pressure on the surface is:

P = \frac{σ ^{2}}{2 ε _{0}}

The force holding the two hemispheres together is this pressure times the cross-sectional area:

F = P \times π R^{2} = \frac{σ ^{2}}{2 ε _{0}} \times π R^{2}

Therefore,

F \propto \frac{1}{ε _{0}} σ^{2} R^{2}

.

The answer is **(A)

\frac{1}{ε _{0}} σ^{2} R^{2}

**.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Single correct option