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Question

(C) The electric field at the surface of a charged spherical shell is $E = \frac{\sigma}{\epsilon_0}$.However, the field acting on the surface charge

Vedclass · Accepted Answer

$\frac{1}{\epsilon_0} \sigma^2 R^2$

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(C) The electric field at the surface of a charged spherical shell is $E = \frac{\sigma}{\epsilon_0}$.However, the field acting on the surface charge itself is due to the rest of the shell, which is $E_{self} = \frac{E}{2} = \frac{\sigma}{2\epsilon_0}$.The force on a small area element $dA$ is $dF = (dq) E_{self} = (\sigma dA) \frac{\sigma}{2\epsilon_0} = \frac{\sigma^2}{2\epsilon_0} dA$.The force $F$ required to hold the two hemispheres together is the net electrostatic repulsive force acting across the cross-sectional area of the hemisphere.The pressure exerted by the electrostatic force is $P = \frac{dF}{dA} = \frac{\sigma^2}{2\epsilon_0}$.The force $F$ is the integral of this pressure over the projected circular area (cross-section) of the hemisphere, which is $\pi R^2$.Thus, $F = P 	imes A_{cross-section} = \frac{\sigma^2}{2\epsilon_0} 	imes \pi R^2$.Therefore, $F \propto \frac{\sigma^2 R^2}{\epsilon_0}$.

$A$ uniformly charged thin spherical shell of radius '$R$' has a uniform surface charge density '$\sigma$'. It is made of two identical hemispherical shells held together by pressing them with a force '$F$' as shown. '$F$' is proportional to [$\epsilon_0$ = permittivity of free space].

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