Question

Eletric felid due to an infinite sheet of charge having surface density isE . Electric feild due to an infinite conducting sheet of same surface density of chargeis

Answer 1

Given :

• Electric field due to an infinite sheet of charge having surface density is E.

To Find :

• Electric Field due to an infinite conducting sheet made up of same surface density of charges.

Solution :

Look at the attached picture :

In First (I) region :

$\implies \sf{E \: = \: \bigg( \dfrac{- \sigma _A}{2 \epsilon _o} \bigg) \: + \: \bigg( \dfrac{- \sigma _B}{2 \epsilon _o}\bigg) } \\ \\ \implies \sf{E \: = \: \dfrac{-1}{2 \epsilon _o} \big[ \sigma _A \: + \: \sigma _B \big] }$

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In second (II) region :

$\implies \sf{E \: = \: \dfrac{\sigma _A}{2 \epsilon _o} \: - \: \dfrac{\sigma _B}{2 \epsilon _o}} \\ \\ \implies \sf{E \: = \: \dfrac{1}{2 \epsilon _o} (\sigma _A \: + \: \sigma _B)}$

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If,

$\sf{\sigma _A \: = \: \sigma, \sigma _B \: = \: - \sigma}$

$\sf{E_I \: = \: 0 \: \: \: \: \: and \: \: \: \: \: \: E_{III} \: = \: 0} \\ \\ \implies \sf{E_{II} \: = \: \dfrac{1}{2 \epsilon _o} \big[ \sigma \: - \: ( - \sigma) \big]} \\ \\ \implies {\boxed{\sf{E_{II} \: = \: \dfrac{\sigma}{\epsilon _o}}}}$

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If density is same then,

$\implies \sf{E \: = \: \dfrac{\sigma}{2 \epsilon _o}} \\ \\ \implies \sf{E \: = \: \dfrac{E_{II}}{2}}$