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[tex] \vec{E_{above}} - \vec{E_{below}} = \frac{\sigma}{\epsilon_{0}} \hat{n} [/tex]

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 [tex] \oint \vec{E} \bullet d\vec{a} = \frac{Q_{enc}}{\epsilon_{0}} [/tex]

and since[tex]Q_{enc} = \sigma A [/tex]

By Symmetry E points up and down

so

[tex] \int \vec{E} \bullet d\vec{a} = 2A |\vec{E}| [/tex]

so [tex] \vec{E} = \frac{\sigma}{2\epsilon_{0}} \hat{n} [/tex]

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 [itex] \vec{E} = \frac{\sigma}{\epsilon_{0}} \hat{n} [/itex]

and te bottom is the negative of that

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

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

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# Homework Help: Electric Field

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