Why is electric field for same surface charge density(σ) is twice in case of a conducting plate or surface (σ/εο) than to a non-conducting sheet (σ/2ε)? One explanation I have found is is because "In the non-conducting surface the entire charge is distributed on both sides of the surface (thus 2 times the area), whereas in the conducting surface, the charge is distributed only on one side." Why is charge in this case distributed to only one side and why not both sides?
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$\begingroup$ More on capacitors and factors of 2: physics.stackexchange.com/q/110480/2451 and links therein. $\endgroup$– Qmechanic ♦Jun 8, 2017 at 7:37
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2 Answers
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In case of nonconducting infinite plane sheet, the charge is distributed in one surface. But in case of conducting plate, charge is distributed in both surfaces. You can think the charged metal plate is equivalent to two infinite parallel plane sheets of similar charges. So the electric field due to conducting plate is twice that of non-conducting sheet
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You can understand this by gauss law. Since in non conducting plate, the flux is via only one surface, whereas in conducting plate, flux is via two of its surfaces which doubles the flux and thus doubles the electric field.
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$\begingroup$ Your answer requires clarification. You should be more elaborate and add in the equations for the two cases. $\endgroup$– MitchellJun 8, 2017 at 14:45