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 sinceQ_{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.