# Electric field outside a capacitor

I know that the electric field outside of a capacitor is zero and I know it is easy to calculate using Gauss's law. We create cylindrical envelope that holds the same amount of charges (of opposite signs) in each plate.

My question is why can't I pick an envelope which include only part of one of the plate? Gauss's law states, specifically, that I can pick any envelope I want.

Note: I have encountered this question a couple of years a go and I got an answer which I was not completely happy with.

• The net field outside an actual capacitor can never be zero. There will always be a fringe field that is too frequently neglected in textbooks. Such idealizations are common, but are also dangerous in that they create misconceptions that are extremely difficult to correct later on. – user11266 Aug 18 '12 at 16:54
• More on capacitors and factors of 2: physics.stackexchange.com/q/110480/2451 and links therein. – Qmechanic May 29 '16 at 9:09
• The fields outside are not zero, but can be approximated as small for two reasons: (1) mechanical forces hold the two "charge sheets" (i.e., capacitor plates here) apart and maintain separation, and (2) there is an external source of work done on the capacitor by some power supply (e.g., a battery or AC motor). Remove (1) and the two "sheets" will begin to oscillate past each other and if we remove (2), then there is no charge on either plate (assuming you did not start with a PS and then remove it after charging up the capacitor), thus no field from the "sheets." – honeste_vivere Jul 11 '18 at 17:29

Outside two infinite parallel plates with opposite charge the electric field is zero, and that can be proved with Gauss's law using any possible Gaussian surface imaginable. However, it might be extremely hard to show if you don't choose the Gaussian surface in a smart way.

The usual way you'd show that the electric field outside an infinite parallel-plate capacitor is zero, is by using the fact (derived using Gauss's law) that the electric field above an infinite plate, lying in the $xy$-plane for example, is given by $$\vec{E}_1=\frac{\sigma}{2\epsilon_0}\hat{k}$$ where $\sigma$ is the surface charge density of the plate. If you now put another plate with opposite charge, i.e. opposite $\sigma$, some distance below or above the first one, then that contributes its own electric field, $$\vec{E}_2=-\frac{\sigma}{2\epsilon_0}\hat{k}$$ in the region above it. Since the electric field obeys the principle of superposition, the net electric field above both plates is zero. The same happens below both plates, while between the plates the electric field is constant and nonzero.

Your way of doing it is a little more tricky, but again gives the same answer. For example, if you choose the Gaussian surface to have an hourglass shape with different radii for the two sides, then indeed the net charge enclosed is not zero. However, when you calculate the total electric flux through that surface, you have to be careful to realize that there is nonzero electric field between the two plates, and therefore there is a nonzero flux through the part of the Gaussian surface that lies between the plates. That flux, of course, has to be accounted for. Assuming that you know the electric field inside the capacitor, $\vec{E}_\text{inside}$, you can do the integral $\oint\vec{E}_\text{inside}\cdot d\vec{A}$ for such a Gaussian surface (it's not that hard actually), and you find that the flux through the part of the surface that lies between the plates is exactly equal to $q_{\text{enclosed}}/\epsilon_0$. Thus, the net flux through the part of the Gaussian surface that lies outside the plates has to be zero, proving, after a little thought, that the electric field outside the capacitor is zero.

The final answer for $\vec{E}$ never depends on the Gaussian surface used, but the way to get to it always does. That's why the Gaussian surface has to be chosen in a smart way, i.e. in a way that makes the calculation of $\oint\vec{E}\cdot d\vec{A}$ easy.

You can pick such a box. However, this box alone does not allow you to determine the electric field. Gauss' law states that the total flux is equal to the charge enclosed times $4\pi$. It doesn't tell you where that flux leaves the box. There could be a lot here, none there, a negative amount at a third place, etc.

In some situations, symmetry allows us to use Gauss' law to find the electric field. A Gaussian box enclosing both plates of a parallel-plate capacitor is symmetric with respect to a reflection through a plane through the middle of the capacitor. It is also symmetric with respect to rotations about an axis perpendicular to the plates (ignoring edge effects). The rotational symmetry lets us say that the electric field first points along the axis perpendicular to the plates. The reflection symmetry tells us that the electric field must be the same through both sides of the box parallel to the plates. That electric field must be zero because the box has no net charge in it.

A box around a single plate does not have the same symmetries. It still has the rotational symmetry, so the electric field must be perpendicular to the plates. However, we don't have the reflection symmetry any more, so the electric field can be different strengths on different sides of the box.

We know there must be net flux through the box because the box encloses net charge. Using the fact that the electric field is zero outside the capacitor, we can deduce the he flux through a box that encloses only one plate is all through the side of the box that's inside the capacitor. Hence, the electric field must be $4\pi\rho$ inside the capacitor.

• Why do we know that the electric field is 0? The intuition is clear but is there something more formal than that? – Yotam Jul 15 '11 at 19:55
• Unless it's a capacitor with infinite plates, the field outside isn't exactly zero. – David Z Jul 15 '11 at 20:56
• It's not only above zero outside but also the path integral over an outer curve is exactly equal to the integral over inner line but of the opposite sign, so the total integral (over a closed path) is equal to zero. – Vladimir Kalitvianski Jul 15 '11 at 21:04
• @David Right. We're neglecting edge effects. @Yolam We know it's zero because we used Gauss' law and the symmetry argument. If you want you could integrate Coulomb's law, but it wouldn't teach you much new. @Vladimir Good point. – Mark Eichenlaub Jul 15 '11 at 22:46
• @Mark: yeah, I was responding to Yolam just to point out that for a real (finite) capacitor, there's no formal way to show that the external field is zero, because is isn't. – David Z Jul 15 '11 at 23:08

The whole premise of this question is wrong. No application of Gauss's law can prove that the electric field outside a capacitor is zero, because it isn't necessarily zero. Here is another situation that satisfies Gauss's law

-----> -     + ----->
-----> -     + ----->
-----> -     + ----->
-----> -     + ----->
-----> -     + ----->


The arrows are electric field lines extending to infinity, while the +'s and -'s are two uniform 2D sheets of charge. You can check that Gauss's law is satisfied for every possible Gaussian surface.

When people say "the electric field is zero outside a capacitor", they are assuming there is no other cause of electric fields besides the capacitor itself. In the example above, if you took the "capacitor" away, there would be a uniform electric field everywhere in space. Why is that field there? Who knows, but it's certainly not because of the capacitor! In other words, they are talking about the electric field created by the capacitor itself.

The proper analysis of a capacitor ASSUMES that the field is zero at infinity and then uses different gaussian surfaces to prove that the field remains zero everywhere outside the capacitor.

Assuming you have only two charges and none outside, you will have the following voltage potential $$\phi$$ as depicted on the left image, and an $$\mathrm{E}$$ intensity as shown on the right image below.

This is simply the the solution of the Laplace Equation

$$\Delta \Phi = - \cfrac{\rho}{\epsilon}\, ,$$

with the boundary condition of two surface charges.

An infinite plate is the only case, where you don't have an electric field outside the capacitor, supposed a nonzero surface charge on the plates, and that both charge densities are of opposite and equal value. Reference of the image:
Solving the Generalized Poisson Equation Using the Finite-Difference Method (FDM) - Scientific Figure on ResearchGate. Available from: https://www.researchgate.net/A-2D-parallel-plate-capacitor-The-top-plate-is-at-a-potential-of-1-0-V-while-the_fig5_228411289 [accessed 11 Jul, 2018]