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Q. A spherical charged conductor has a as the surface density of charge. The electric field on its surface is $E$. If the radius of the sphere is doubled keeping the surface density of charge unchanged, what will be the electric field on the surface of the new sphere ?

Electrostatic Potential and Capacitance

Solution:

$E = \frac{1}{4 \, \pi \, \varepsilon_0} \frac{Q}{R^2} = \frac{1}{4 \, \pi \, \varepsilon_0} \, \frac{4 \, \pi \, R^2 \, \sigma}{R^2} = \frac{\sigma}{\varepsilon_0}$
It is independent of radius and depends on $\sigma$