How Is the Electric Field Difference Across a Charged Surface Determined?

[madafo3435]

Homework Statement

I am studying the book Electricity and Magnetism by Edward M. Purcell. I resort to the forum because I have problems in understanding where the equation (1.45) that I indicate at the end of my attached images comes from.

Relevant Equations

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I understand that this difference is valid for E = 0 and E2 = σ / Ɛ0, but Purcell covers a more general case, and I don't see how this difference is fulfilled in other cases.

I appreciate the help you can give me.

 

[etotheipi]

Take a surface of some charge density, and of any general shape. At some point on the surface, construct a very tiny Gaussian pillbox that passes through the surface. At all points on the surface the field due to the surface is orthogonal to the surface, so the flux through the sides of the pillbox approaches zero as we make the pillbox smaller. The net flux out of the pillbox, if the electric fields on either side are $\vec{E}_1$ and $\vec{E}_2$, will be$$\Phi = \vec{E}_2 \cdot \hat{n} \delta S + \vec{E}_1 \cdot (-\hat{n}) \delta S = (\vec{E}_2 - \vec{E}_1) \cdot \hat{n} \delta S = (E_{n,1} - E_{n,2}) \delta S = \frac{q}{\varepsilon_0} \implies E_{n,1} - E_{n,2} = \frac{\sigma}{\varepsilon_0}$$since the unit outward normals, $\hat{n}$ and $-\hat{n}$, to the two opposite faces of the pillbox will be oppositely directed.

 

[madafo3435]

Take a surface of some charge density, and of any general shape. At some point on the surface, construct a very tiny Gaussian pillbox that passes through the surface. At all points on the surface the field due to the surface is orthogonal to the surface, so the flux through the sides of the pillbox approaches zero as we make the pillbox smaller. The net flux out of the surface, if the electric fields on either side are $\vec{E}_1$ and $\vec{E}_2$, will be$$\Phi = \vec{E}_2 \cdot \hat{n} \delta S + \vec{E}_1 \cdot (-\hat{n}) \delta S = (\vec{E}_2 - \vec{E}_1) \cdot \hat{n} \delta S = (E_{n,1} - E_{n,2}) \delta S = \frac{q}{\varepsilon_0} \implies E_{n,1} - E_{n,2} = \frac{\sigma}{\varepsilon_0}$$since the unit outward normals, $\hat{n}$ and $-\hat{n}$, to the two opposite faces of the pillbox will be oppositely directed.

His explanation was very clear, now I understand. Thank you very much