I've read over several other discussions on the topic of "electric field of a sheet of charge vs electric field of a charged plate", on this very site, but I still haven't gotten the answer to my question. And what's made me so curious is a combination of a tutorial exercise from a book called "Tutorials in Introductory Physics" about capacitance that talked about how, for a charged plate, the charges on the two facings of the plate, through superposition, cancel their fields in the center BUT their fields extend out to either side - BOTH facings' charge distributions extending out BOTH sides, keep in mind.

But then I'm working on some college board questions for my students, and I come across this problem, which I'll summarize: We take an isolated plate, with face areas A1 and A1, and give it a charge +Q. We then bring it near another plate with a charge -Q to make a capacitor, and they want to know the electric field strength a short distance away from the + plate after vs before. They say it's 4 times as strong, using the logic that all the charge that was split between two facings with area A1 now ends up on one side, doubling charge density and doubling the electric field, then you bring in another plate which adds to the electric field because their fields point the same way (basic inside of capacitor thinking), and voila - double double for 4x the field.

I have a problem.

Does moving all the +Q charge from being distributed around both areas to just being on one ACTUALLY double the electric field? My math says no.

I hate to go into the gauss's law stuff AGAIN, but here goes:

Let's look at the plate. It has a charge of +Q. on one side there's 1/2 Q and the other there's also 1/2 Q. Or you can say that σ1, as I'll call it, is Q/2A1, since it's distributed across both faces. Fine. Doesn't that mean then that on one side of the plate, there's a superposition of electric field from a sheet with charge density σ1 AND a second sheet with charge density σ1? I know they cancel in the middle, but outside there should be superposition, right?

And if so, then I feel like we should get the same answer using gauss's law whether we draw our gaussian surface to have part inside the conductor or to have both sides of our gaussian surface "flanking" the plate.

Gaussian surface inside the conductor: qenclosed is σ1A, over ε, and is equal to EA, A being the area of our surface not the plate. This cancels and gives us σ1/ε. Since σ1 is Q/2A, we could write this as Q/2Aε, but it's really σ1/ε which is consistent with the idea of the electric field of a conductor. The other side's charge isn't included in the gaussian surface, so the net flux it contributes is zero.

Gaussian surface ACROSS the conductor: Now the amount of charge we enclose is doubled from before, so qenclosed is 2σ1A. Put that over ε, and this is equal to EA + E*A, since now the electric fields extend out both sides and pass through two of the areas of our gaussian surface. This gives 2σ1A/ε=2EA, the 2 coefficients and area of gaussian surface A cancel, leaving us with, once again, σ1/ε. We get the same result as before, because we're still trying to solve for the electric field of a conductor at equilibrium.

Now, let's squish it down. If we collapse all the charges together, either because we're imagining the plate as a sheet OR, like in the problem, we've brought another -Q charged plate nearby so that all the +Q charge gathers on only ONE surface of our plate, this doubles the charge density. I'll now just call that σ instead of σ1, and this new charge density σ is 2*σ1. Since σ1 was equal to Q/2A1, this makes the new charge density equal to Q/A1. Pretty standard.

This is where things get tricky again. We KNOW already that for our capacitor, we can model the positive plate as a sheet of charge. It's all over the internet and in our textbooks. So the field from that plate should be σ/2ε, which adds together with the field from the -Q plate, which is also σ/2ε, and gives us σ/ε. σ is 2*σ1, so this electric field is twice as strong as before - not 4 times! I've also seen this done modeling the plate of a capacitor as a conductor. In those cases, because of gauss's law in use with a conductor, we get an electric field σ/ε only. This is also only 2 times as strong as our previous case.

But, curiously, I never see anything BOTH A) treat the positive plate as a conductor, which gives the electric field σ/ε, AND add it to the electric field from the negative plate, which would also be σ/ε, for a net result of 2σ/ε. That would, if done, explain the answer of 4x as strong of an electric field but I never see that. It's either "look it's a sheet, σ/2ε + σ/2ε = σ/ε" or "it's a capacitor, one plate is a conductor, its field is σ/ε and we're done", never "and let's add the other one now, too". Doing so would yield different results based on whether you treated the plates as sheets of charge or as conductors, so I just don't see this done.

The college board question just says that condensing the charge to one side doubles the density and therefore doubles the electric field prior to it doubling again by the addition of the other plate, but I've shown this not to be the case (I think), so is college board wrong and the electric field inside the capacitor is only 2x as strong as the +Q plate in isolation? Or have I messed up an application of dividing by 2 or multiplying by 1/2 somewhere? Or do I have a fundamental issue somewhere with how I've applied Gauss's law? Can you not have your gaussian surface extend from just outside one side of a conductor to just outside the other? If not, why can't you?

Does the electric field from the charges on one side of a charged plate add in superposition to the electric field on the other side, cancelling in the middle? If not, why not? Superposition is the only reason the electric field cancels in the middle, so why would superposition "stop" or "not exist" outside the outer facings?

My poor students have tolerated my muddling through this long enough. Please help me figure this out!

  • 2
    $\begingroup$ Your question is very long and not easy to get through, plus you claim your math shows something is untrue and you provide zero math to show that. You really need to focus the question down so it's far easier to read and digest. $\endgroup$
    – Triatticus
    Mar 8 at 1:45
  • $\begingroup$ I showed the math, please check again. I show that the field around the charged plate is the same whether it's treated as two sheets of charge, a charged conductor, or a single sheet with twice the charge density, and does not double the electric field like the CB question claims. $\endgroup$ Mar 8 at 17:10

1 Answer 1


According to Bill N’s answer to this question, you are quite right.link to PSE question The charges on each side of the plate contribute equally (add) to the field outside of the plate. So for a plate with charge density $\sigma$ on each side, he gives the total field is $\frac{\sigma}{\epsilon_0} \hat n$
Or simply consider two charges on a line. Exactly between the charges their fields do cancel, but to either side of the charges the fields add.

  • $\begingroup$ So then moving all the charge from one side to the other so that it's all together with twice the charge density shouldn't double the electric field, right? The field strength will still be the same, as long as it's the same amount of charge? $\endgroup$ Mar 8 at 17:13
  • $\begingroup$ That's how I understand it. It looks to me as if in the question they assumed that the charge on one side of the plate did not extend to the other, perhaps because they thought that a field does not cross a conducting sheet? $\endgroup$
    – Rich
    Mar 8 at 19:21
  • $\begingroup$ I do agree with triatticus that your presentation was a bit of a muddle. It really helps to use mathjax (math.meta.stackexchange.com/questions/5020/…) which only takes a short time use get used to. $\endgroup$
    – Rich
    Mar 8 at 19:36

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