# Electric field due to nonconducting plastic sheets [closed]

Two very large, nonconducting plastic sheets, each 10.0 cm thick, carry uniform charge densities $\sigma_1, \sigma_2, \sigma_3$ and $\sigma_4$ on their surfaces (the four surfaces are in the following order $\sigma_1, \sigma_2, \sigma_3$ and $\sigma_4$ going from left to right). These surface charge densities have the values $\sigma_1 =-6, \sigma_2 = +5, \sigma_3 = +2$ and $\sigma_4 = +4$ all in C/(m*m).

A) Use Gauss's law to find the magnitude of the electric field at the point A, 5.00 cm from the left face of the left-hand sheet.

B) Find the magnitude of the electric field at the point B, from the inner surface of the right-hand sheet.

C) Find the magnitude of the electric field at the point C, in the middle of the right-hand sheet.

So in the book the author derives an equation for this kind of situation, and there is actually an example in the text, but the example is for a plate that has only one charge and not two like this one. Anyways, the equation is $E=\frac{\sigma}{2\epsilon_0}$.

I got question A) right by just adding up the electric field due to all the charges, namely $E_{net}=\frac{\sigma_2}{2\epsilon_0} +\frac{\sigma_3}{2\epsilon_0}+\frac{\sigma_4}{2\epsilon_0}-\frac{\sigma_1}{2\epsilon_0}$, but I am not sure about B) and C), would they be the same? When I drew the electric field vectors, going towards the negative side and away from the positive one, I couldn't really figure out what direction the net vector would be pointing to.

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## closed as too localized by Qmechanic♦Feb 15 '13 at 15:32

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Hi Daniel, and welcome to Physics Stack Exchange! Any chance you can post an image of the electric field vectors you drew, and showing the positioning of points B and C? – David Z Jan 27 '12 at 23:49
It won't let me, it says I need to have at reputation of at least 10 to post images. – Hiro Jan 28 '12 at 0:14
Oh, right, sorry about that. If you can get the image online somewhere, just post the URL and someone with higher reputation can edit it into your post for you. – David Z Jan 28 '12 at 0:23

The key to solving this question is applying Gauss' Law through drawing Gaussian surfaces. The example in your book probably reached the result you give by drawing a cylinder perpendicularly through a charged sheet. If not, see the explanation here under the section "Field of an Infinite Plane Sheet of Charge". You need to understand how exactly this gives you the result $E=\frac{\sigma}{2\epsilon_0}$