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We know for an infinite plane sheet, electric field from the sheet is given by: $$ E = \frac{\sigma}{2\epsilon_0} \hat n$$

Therefore potential is given by $$ - \frac{ \partial V}{\partial n} = \frac{\sigma}{2 \epsilon_0} $$

However, in Griffiths, page 125, 4th edition, section 2.2 on potentials, it says:

$$ \sigma = - \epsilon_0 \frac{\partial V}{\partial n} $$

Where did I go wrong?

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  • $\begingroup$ If the sheet is thick, then it will have two surfaces, and the field from both of them add to give electric field $E = \frac{\sigma}{\epsilon_{0}}$ $\endgroup$ – Shine kk Apr 29 at 5:56
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On page 125 of the 4th edition (I'm reading the international edition) of Griffiths that equation is not attributed to an infinite plane sheet but rather to the surface of a conductor, and has to do with the discontinuity in the field. He explains in more detail on page 88 and page 103.

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It could be that he might have been talking about two infinite sheets (metal plates) of charge densities $+\sigma$ and $-\sigma$ (capacitor plates or something) - with $E=\sigma/\epsilon_0$.

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In addition to the above ; the discontinuity at the plane is twice the absolute field value on each side, as they have opposite directions.

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