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Please kindly refer to page 88 in the link below

Click here

Extracted image from "Click here" For a sheet with surface charge $\sigma = \sigma_{f}+\sigma_{b}$, the electric flux through the gaussian pill box of area A can be expressed with Gauss's law:

$\oint_{S}\vec{E}\cdot d\vec{a}=\frac{Q_{enc}}{\epsilon_{0} }=\frac{\sigma A}{\epsilon _{0}}$

The pill box has an area vector $\vec{A}=\pm \hat{z}A$: "-" if the area A is facing in the negative z direction and "+" if the area A is facing in the positive z direction.

We know that the sheet produces its own electric field due to the surface charge $\sigma$. This electric field is in the +z direction above the sheet and in the -z direction below the sheet.

Now, there is, without Griffith made explicitly clear, an external electric field below the charge sheet. This electric field is in the z direction and the field lines from this external electric field 'pass' through the charge sheet.

Here is where things gets confusing:

Griffith asserts that$ E^{\perp}_{above}-E^{\perp}_{below}=\frac{\sigma}{\epsilon _{0}}$

Is this $E^{\perp}_{above}, E^{\perp}_{below}$ a result due to the external electric field or is it a net electric field due to the electric field from the sheet and the external electric field? While I'm not sure, I am inclined to say this cannot be the case since $\frac{\sigma}{\epsilon _{0}}$ is due to the charge enclosed IN the gaussian pill box.

Someone please shed some light.

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The $E^{\perp}_{above} and E^{\perp}_{below}$ refer to the perpendicular components of the total electric field. This includes thus both the external electric field and the field generated by the charge on the sheet.

If we write this out: $$E^{\perp}_{above} - E^{\perp}_{below}=\frac{\sigma}{\epsilon_0}$$ $$E^{\perp}_{above,sheet} + E^{\perp}_{above,ext} - E^{\perp}_{below,sheet} - E^{\perp}_{below,ext}=\frac{\sigma}{\epsilon_0}$$ The external field will be identical above and below the sheet thus $$E^{\perp}_{above,sheet} - E^{\perp}_{below,sheet} =\frac{\sigma}{\epsilon_0}$$

Due to the symmetry of the problem we can assume that $E^{\perp}_{below,sheet}$ and $E^{\perp}_{above,sheet}$ are equal in size, but opposite in direction. This teaches us that a charged sheet creates an electric field of magnitude $$E_{sheet}=\frac{\sigma}{2\epsilon_0}$$ pointing away from the sheet

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  • $\begingroup$ Based on what has been shown by Griffiths together with what you've said, can it then be concluded that $E^{\perp}_{below}$ is the net electric field below the sheet- 'surviving external electric field after superposition? $\endgroup$ – Physkid Aug 31 '16 at 12:36
  • $\begingroup$ @Physkid what do you mean by net electric field? And by surviving external electric field after superposition? $\endgroup$ – Crimson Aug 31 '16 at 12:43
  • $\begingroup$ I meant by the superposition of the component of the external electric field and component of the electric field due to the surface charge below the sheet. $\endgroup$ – Physkid Aug 31 '16 at 12:45
  • $\begingroup$ then yes, that is exactly what it means $\endgroup$ – Crimson Aug 31 '16 at 12:49
  • $\begingroup$ And since $E^{\perp}_{above}-E^{\perp}_{below}=\frac{\sigma}{\epsilon _{0}}$ results, it can be concluded that it must be the case that external electric field is killed off. Is this correct? Now, is there a proof to actually show that the external field is killed? Or is this a result of experiment? $\endgroup$ – Physkid Aug 31 '16 at 12:51

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