Signs in application of Gauss's law to a sheet of charge In Purcell's Electricity and Magnetism, page 52, the author said that in the following situation:

[Where we have a slab of positive charge with uniform density $\rho$. $\,\bf{E_{1}}\,$ is the $E$-field to the left of the slab and $\,\bf{E_{2}}\,$ the one to its right. We define $\,\sigma=\rho\Delta r\,$ as the surface charge density on each layer (which we assume to be constant)]

The condition imposed by Gauss's law, for given $\sigma, $ is 
  $$E_{2}-E_{1}=4\pi\sigma$$

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Shouldn't it be $\,E_{2}\, \textbf{+} \,E_{1}\,$ instead of $\,E_{2}-E_{1}$ ?
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P.S. We are using cgs units in the quoted formula.
 A: I don't have the book so I can't follow the derivation exactly, but I think this comes down to the orientation of the unit normal vector for the Gaussian surface being used. Conventionally, the unit normal vector always points outwards from the surface, and so contribution to the flux integral of $E_1$ and $E_2$ shall have opposite signs.
Another thought on this is, it makes sense that the presence of a surface should affect the difference between the fields on either side, rather than their sum. The sum of the fields would depend on external factors, perhaps like the magnitude of charge used to produce them. The difference between fields across a boundary, however, should only really depend on what's happening on the boundary (i.e. the charge density, here).
A: The field on either side of the sheet, found using Gauss's law, is 
$$\mathbf E=2\pi\sigma\ \hat n$$
where $\hat n$ points away from the sheet of charge. Therefore, if we take the sheet to be perpendicular to the $x$-axis we have
$$\mathbf E_1=-2\pi\sigma\ \hat x$$
$$\mathbf E_2=2\pi\sigma\ \hat x$$
Therefore:
$$||\mathbf E_2-\mathbf E_1||=4\pi\sigma$$
So your book is correct. This is what is typically used to show how we have a discontinuity in the field when moving across a sheet of charge.
