# Surface charge density formula

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?

• If the sheet is thick, then it will have two surfaces, and the field from both of them add to give electric field $E = \frac{\sigma}{\epsilon_{0}}$ – Shine kk Apr 29 at 5:56

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 $$E=\sigma/\epsilon_0$$.