The magnitude of electric field on the surfac | vedclass

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Question

(A) The electric field just outside a charged conductor is given by $E = \sigma / \epsilon_0$,where $\sigma$ is the surface charge density.This field

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$E/2$

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(A) The electric field just outside a charged conductor is given by $E = \sigma / \epsilon_0$,where $\sigma$ is the surface charge density.This field is the result of the superposition of the field due to the small patch where the hole is made $(E_{patch})$ and the field due to the rest of the shell $(E_{rest})$.At a point just outside the surface,$E_{patch} + E_{rest} = E = \sigma / \epsilon_0$.Since the field due to a small patch is $E_{patch} = \sigma / (2\epsilon_0)$,we have $E_{rest} = E - E_{patch} = \sigma / (2\epsilon_0) = E/2$.At a point just inside the surface,the fields are $E_{patch} - E_{rest} = \sigma / (2\epsilon_0) - \sigma / (2\epsilon_0) = 0$.When a hole is made,the field inside the hole is due to the rest of the shell $(E_{rest})$,which is $E/2$.

The magnitude of electric field on the surface of a uniformly charged metallic spherical shell is $E$. If a hole is made in it using an insulating device,then the magnitude of electric field in the hole will be

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