While I was studying about capacitors I realized that charge on the outer surface of the plates is zero when both the plates are given equal and opposite charges. Now mathematically, this realization came to using the fact that the net field inside any point inside the plate must be zero.

But practically, looking at the electrons' behavior, does this mean that all the charges come together because they are kept so close? Would the plate have equal charges have throughout their respective surfaces when kept far apart but not at infinity?

I tried to visualize this myself, but I am not confident with my thought process. First of all, I imagined that in the above situation, the positive charges, or the "voids" grouped together on one surface only because they were kept so close, and similar happened with the electron. Following this, I now pictured two plates, one with charge Q and other with none. They are brought near and according to my thought process, it must be Q and -Q, as due the small distance, the "gaps" and the electrons all should rush towards the gaps. So the Q on the originally charged plate attracts all the -Q equivalent electrons, leaving a "void" of +Q on the other surface. But when I tried using the fact that the net electric field inside either of the plates must be zero, I was instead left with charges +Q/2 and -Q/2 and +Q/2 on the outer surface of both the plates.

So where was the flaw in my thinking? I just assumed that the +Q would attract all the electrons worth -Q. But why does this not happen. This means that my thinking in the very first question, though gave me the right answer, but apparently was not correct.

I was looking for an explanation but was unsuccessful in it. Please note that I have already derived what charges should appear on the plates and I am aware of that. I am just looking for a more 'intuitive' or 'imaginable' reason as to why the electrons behave like this.

As a brief example, consider a point charge q, which is brought near a neutral, spherical conducting sphere. So the sphere must gain -q near the point charge and +q on the other side. Why doesn't the same happen with the metal plates?

Also, does the same happen with concentric conducting shells? Like, if the inner shell is charged to some charge, would the same charge appear on both sides of the outer shell?


3 Answers 3


It is approximately true that:

that charge on the outer surface of the plates is zero when both the plates are given equal and opposite charges.

We will get to why it is only approximately true, and not exactly true in a moment.

does this mean that all the charges come together because they [the plates of the capacitor, I am assuming] are kept so close?


Imagine that instead of two closely spaced parallel plates, you had two oppositely charged spheres (and nothing else in the universe). The electric field would be present throughout space, and there would be field lines terminating at every point on the surface of each sphere. (Assuming charges are "continuous" rather than discrete).

Now gradually flatten the spheres into thin plates, and move them closer together. As the "spheres" flatten and move closer together, the electrons and holes will re-arrange toward the configuration found in a thin plate capacitor. But unless the capacitor plates are infinite, or the space between them infinitessimal, there will still be charges everywhere on the entire surface of the plates, both the side facing toward the other plate and the side facing away. Further, the field between the plates will not be uniform, but will bulge away from an axis passing perpendicularly through the plates.

Quantitatively, as the plates become flatter and closer, the field on the "outside" surfaces of the plates becomes smaller and smaller, and the bulging of the field between the plates becomes smaller and smaller except at the very edges of the plates, where it is called fringe effects. One can estimate this quantitative effect by the length of the field lines. Field lines going from the "outer" surface around the capacitor to the outer surface of the other plate are much much longer than the field lines going in-between the plates. So the field strength "outside" the capacitor will be much, much, smaller. So small, that we normally treat it as negligible. However, by using the method of a thought experiment where one is continuously distorting the "plates" from spheres into close flat plates, one can gain an intuitive understanding how the electrons and fields in a capacitor are arranged.

  • $\begingroup$ But suppose we have unequal charges on the plates say Q and 0. Then why don't I end up with charges -Q and +Q on the other plate and instead have -Q/2 and +Q/2? Why wouldn't the complete Q charge on the originally charged plate not induce complete charge on the second plate? $\endgroup$
    – Adi
    Aug 2, 2021 at 15:08
  • $\begingroup$ For instance, when we bring a charged body with charge Q, say near a neutral conductor, the magnitude of charges induced on the conductor would be Q. So the original charge is induced completely. Then why, in case of the plates, does this not happen? $\endgroup$
    – Adi
    Aug 2, 2021 at 15:22

Typically parallel plate capacitors are idealized somewhat. The field between them is considered uniform. Normally this is a very good approximation. But it is strictly true if the plates are infinite. Near the edges, the field is not uniform.

Normally the plates are very close together. The width is much bigger than the separation. You are wondering what happens if you make the separation bigger.

If you move the plates very far apart, so the width is small compared to the separation, the capacitor resembles to separated point charges more than a capacitor. The field is not uniform. The charges would spread out all over the plates.

If you truly did have infinite plates, The field between them would stay uniform, no matter how far they were separated. The charges would stay on the inside.

Intuitively, the charges are on the inside because the opposite charges on the two plates attract each other.

If the plates are separated and the field stays uniform, the field is just as strong. The electric forces of attraction are just as strong. No surprise that the charges stay on the inside.

Note that the energy and capacitance do change if you separate the plates. It takes work to separate two plates that attract each other. It increases the potential energy.

$$C = q/V = q/(potential \space energy \space per \space electron)$$

Increasing the separation decreases the capacitance.


I'll just look at the interesting case of a plate with charge $Q$ facing a neutral plate.

In the plate dimension >> plate separation, uniform field approximation, the charges on the plate surfaces are:

$+Q/2$ on each surface of the charged plate. $±Q/2$ on the surfaces of the neutral plate, with $-Q/2$ on the inner (facing) surface and $+Q/2$ on the outer surface.

Taking account of the charges on all 4 surfaces, and the fields that they give rise to, you'll find that the resultant field inside each plate is zero. [It's useful to bear in mind that, because of the uniform field approximation, the field outside a plate is the same as if the net charge on the plate were evenly distributed over its surfaces. One consequence is that, even though the neutral plate has charges ±Q induced on its surfaces, it makes no field at points outside itself (provided these are close compared with the plate's linear dimensions).]

How does the actual surface charge distribution come about (given the overall charges of $+Q$ and 0 on the plates)? By movement of free electrons within the metal of the plates. If there is a net electric field within the metal the free electrons will move in the direction of the field, though frequently colliding with the lattice. Suppose that the neutral plate was put in position near the positive plate and had, at one instant, charges of $±Q/4$ on its surfaces. There would then be, it is easy to see, a resultant electric field inside the neutral plate so the free electrons would be moving. Movement would stop when the charges on its surfaces were $±Q/2$, as the electric field inside it would be zero.

  • $\begingroup$ I am aware of that and have deduced a formula that on the inner surfaces, the magnitude of the charges will be (Q1 - Q2)2 using the principle you just mentioned. I rather wanted to know if there was a more intuitive explanation that why wouldn't electrons of a total charge -Q on the inner surface of the neutral plate instead of -Q/2 $\endgroup$
    – Adi
    Aug 2, 2021 at 15:15
  • $\begingroup$ Sorry. But the zero-field-in-a-conductor-at-equilibrium condition does have an intuitive interpretation. If there were a resultant field the free electrons would move under its influence. They stop moving only when they are distributed in such a way that they provide a field, which together with any field from outside (in this case the surfaces of the positive plate) make a resultant field of zero! I've added a paragraph to my answer to try to bring this out. $\endgroup$ Aug 2, 2021 at 17:52
  • $\begingroup$ Thanks. It is quite clear now. Before, I thought that the "field zero inside conductor" was merely used as a tool to calculate charges on the plate. But now is it correct to conclude that BECAUSE field must be zero, such a charge distribution? $\endgroup$
    – Adi
    Aug 2, 2021 at 18:57
  • $\begingroup$ Yes, because for equilibrium the field must be zero, the charge distribution must be so in electrostatics. $\endgroup$ Aug 2, 2021 at 19:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.