- #1
jegues
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Homework Statement
Using Gauss' law, derive the expression for the electric field intensity vector of an infinite sheet of charge in free space.
Homework Equations
The Attempt at a Solution
See figure attached for their solution and the figure that goes with the problem.
What do they mean by planar symmetry? Is it simply that the whole surface lies in a plane and is symmetric?
Also it's not obvious to me as to how they developped their answer.
Using Gauss's law,
[itex]\oint_{S} \vec{E}\cdot\hat{n}dS = \frac{Q_{enclosed}}{\epsilon_{0}}[/itex]
First we note that the electric field vectors in space due to the sheet are normal to the sheet.
Now if I enclose the sheet in a rectangular cube, the electric field vectors will be pointing parallel to the normal vectors on the top and bottom of the rectangular cube.
Thus,
[itex]E\oint_{S}dS = \frac{Q_{enclosed}}{\epsilon_{0}} [/itex]
The surface area of the rectangular box, [itex]\oint_{S}dS[/itex] should simply be 2(lh) + 2(lw) + 2(wh), where l=length, w=width, h=height.
I don't see this approaching the answer they provide.
Any ideas on where I'm going wrong, or how I'm misinterpreting their solution?
Thanks again!