Electric Field and Gaussian Pillbox

[stunner5000pt]

check whether this result of this problem is consistent with this statement
$$\vec{E_{above}} - \vec{E_{below}} = \frac{\sigma}{\epsilon_{0}} \hat{n}$$

an infinite plan carries a uniform surface charge sigma. Find its electric field
Solution:
Draw a Gaussian Pillbox extending above and below the plane. Then
since $$\oint \vec{E} \bullet d\vec{a} = \frac{Q_{enc}}{\epsilon_{0}}$$

and since$$Q_{enc} = \sigma A$$

By Symmetry E points up and down

so
$$\int \vec{E} \bullet d\vec{a} = 2A |\vec{E}|$$

so $$\vec{E} = \frac{\sigma}{2\epsilon_{0}} \hat{n}$$

Now to tackle the question

Well for an infinite sheet for each side the Electric field points normal to the sheet, right?
SO the electric field for the top is $\vec{E} = \frac{\sigma}{\epsilon_{0}} \hat{n}$
and te bottom is the negative of that

So when you add those two together you get $$\vec{E} = \frac{\sigma}{\epsilon_{0}} \hat{n}$$

and this is consistent with statement. Easy enough. i just want to know whether this kind of 'proof' for the statement is satisfactory.

 

[pseudovector]

yeah that's right.