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Suppose I have two metal plates in a vacuum and I give this system some electric charge,the charge would distribute itself according to Gauss law on both the inner and outer walls of both plates...but if I were to hook the plates to a battery and convert this system into a capacitor ...all the charge would be on the facing surfaces and none on the outer walls ..why is this so??

I had seen a question in Concepts of Physics by HC Vermain which a system of two plates was given and one plate was charged with +Q and they said to find the potential difference between the two plates but in applying Q=CV they considered only the facing side's charges and my teacher said that this is because in a capacitor charges only reside on the inner walls and hence we are not considering the charges on the outer walls in consideration and when we hook the capacitor to a battery there will be no charge remaining on the outer walls

What is the precise math and corresponding equations behind it? Is there any explanation based on electric field and potential?

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  • $\begingroup$ Such a thing did not happen when the plates weren't hooked up to the battery?...Also can u describe this mathematically? $\endgroup$ – Schwarz Kugelblitz Aug 14 '19 at 14:49
  • $\begingroup$ Who says there isn't any charge on the outer walls? $\endgroup$ – Michael Seifert Aug 14 '19 at 15:20
  • $\begingroup$ At the star when you give the two plates some charge is it of the same sign on the two plates? Further, is the magnitude of the charge the same on both plates? $\endgroup$ – Farcher Aug 14 '19 at 15:20
  • $\begingroup$ @MichaelSeifert my teacher and a bunch of books and questions I saw say so $\endgroup$ – Schwarz Kugelblitz Aug 14 '19 at 15:36
  • $\begingroup$ @Farcher let us consider that I am giving a charge +Q to one of the plates and no charge to the other $\endgroup$ – Schwarz Kugelblitz Aug 14 '19 at 15:37
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One must clearly distinguish between the case of a capacitor free floating in vacuum and a capacitor connected to a circuit. Once you connect your capacitor to a circuit, any charge you had previously is not relevant, and the characteristics of the circuit determine the charge on the capacitor.

If the plates are in free space then the outer face can indeed be charged. For example, if you put +Q charge on plate 1, a charge of -Q is induced on the inner face of plate 2, but since plate 2 is neutral, a charge of +Q is formed on the outer face of plate 2.

On the other hand, if your plates are connected to an electrical circuit, there can be no charge on the outer face of the plates because these charges on the edge of the plate would immediately be neutralized by charges flowing from/to the circuit. The charges on the outer face are only there because they are pushed to the edge by electrostatic forces, and connecting the plate to a circuit allows them an escape route. The circuit, being a reservoir of charge, can supply or remove any charge needed to neutralize the outer faces.

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It happens because of the way charges are arranged on the plates.

For the case of infinite plate capacitors, we can assume uniform charge density on all the plates of the blocks. Let's say the surface charge density is $\sigma$ and $-\sigma$ respectively.

Now let's assume the charges are spread out on both sides of the capacitor plates.

From Gauss law, we can find that the absolute value of electric field intensity at any point P due to any of the four plates is

$$|\dfrac\sigma{2\epsilon_0}|$$

Now if we consider a very small region of some little volume in the bulk of one of the plates (say the positive one), we can see that the net E-field there isn't zero. Its $$\dfrac\sigma{2\epsilon_0}+\dfrac\sigma{2\epsilon_0}+\dfrac\sigma{2\epsilon_0}-\dfrac\sigma{2\epsilon_0}$$ $$=\dfrac\sigma{\epsilon_0}$$

Now, if we look at the force intensity experienced by the charges on the leftmost face, it is $\dfrac\sigma{2\epsilon_0}$

(that's because now we've considered the intensity only due to the remaining three charge distributions)

Clearly it isn't zero.

Hence those charges on the farther plate of the block will move under the influence of this field and end up on the inner face.

We can also show this for the negative charges.

enter image description here

In the diagram I've marked the field lines. At a point what matters is the net field or the sum of all the fields. We can see that the net field in that tiny volume is towards right.

I didn't draw the field lines for all the charges as it would make the diagram more clumsy.

However in case of a capacitor which is hooked up to a battery it's not true that the outer surfaces won't have any charge because after connecting the battery when charges aren't moving anymore we say the plate and the terminal to which it's connected are at the same potential. So the one of the terminals will be at the same potential as the positive terminal of the battery and the other one will at the same potential as the negative terminal.

Even if it were not the case, the battery would soon distribute the charges such that the plates and terminals are at the same potential.

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