The question: 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, as shown in the following figure. These surface charge densities have the values $\sigma_1= -6.30$ $\mu \text{C}/m^2$ , $\sigma_2= 5.00$ $\mu \text{C}/m^2$, $\sigma_3= 2.10$ $\mu \text{C}/m^2$, and $\sigma_4= 4.00$ $\mu \text{C}/m^2$. Use Gauss's law to find the magnitude and direction of the electric field at the following points, far from the edges of these sheets.

enter image description here

Attempt: So, using Guass' law, I attempt to place the charge densities into a surface and use $E = \frac{\sigma}{2\epsilon_0}$, representing the sheets as infinite thin sheets.

For A, I have a surface encompassing every $\sigma$. $\sigma_1$ is negative, so it goes into the surface, and the other three are inside the surface and positive, so they are going from inside to outside the surface. Hence, for A, it should be $\frac{\sigma_2 + \sigma_3 + \sigma_4 - \sigma_1}{2\epsilon_0}$.

For B, I have a surface encompassing $\sigma_1, \sigma_2$ and one encompassing $\sigma_3, \sigma_4$. Well, again $\sigma_1$ goes into the surface, and the other three go out of their surfaces, so it would be the same answer as for A, but apparently this isn't right. I would use the same methodology for C, but it would be wrong. I don't understand how to approach this.

Edit: I have read solutions to this elsewhere, but they aren't in-depth enough and aren't satisfactory.

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  • $\begingroup$ When you consider $B$, pay attention to the direction of the fields from each surface charge. $\endgroup$ – garyp Apr 11 '15 at 16:56
  • $\begingroup$ Why has this question been resurrected by "Community" when an almost identical question which has hints towards the answers been [closed] by @Qmechanic? physics.stackexchange.com/q/20109 $\endgroup$ – Farcher Mar 14 '16 at 7:06
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    $\begingroup$ @Farcher: The community bot regularly bumps questions with no upvoted answer, cf. this meta post. As for why the other question was closed and this one was not, I'd guess it's simply because no one ever cast a close vote on this one. $\endgroup$ – ACuriousMind Mar 14 '16 at 10:51

I am unable to add comments, so I give a few hints here

  1. Look at the derivation of the formula you used for computing field - check its assumptions and scenario carefully - it is slightly different from what you have here
  2. Use the same method as that in the derivation, but, be careful about the values of the field strengths, and you can get the answer.
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  • $\begingroup$ Why can't you add comments you have enough rep? $\endgroup$ – Gonenc Jun 16 '15 at 12:16
  • $\begingroup$ Obviously, this answer was written when I didn't have enough rep. :D $\endgroup$ – tpb261 Jun 16 '15 at 12:55
  • $\begingroup$ Shoot the answer is from 2014! $\endgroup$ – Gonenc Jun 16 '15 at 13:12

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