0
$\begingroup$

I recently studied about earthing of 2 parallel conducting plates.One is given a chrage q and other is kept at a distance d and charge is induced on it. Now it is earthed. Now it was given that the charge on outside of both the plates will be 0 and thus there will be no net electric field outside the plates and potential of the earthed plate will become 0. But i came across a paradox. If the potential of the earthed plate is 0 the potential of other plate should also be 0 as there is no electric field outside it and also it has simillar charge distribution but there exists electric field between the two plates so potential should vary as E*d . Can someone pls explain what am i missing enter image description here

plates are considered as infinte

$\endgroup$
1
  • $\begingroup$ I wonder whether your question has much to do with earthing. Surely any way of putting equal and opposite charges on the plates, for example connecting a battery across them, would give you the set-up needed for your question. $\endgroup$ Jun 15, 2023 at 17:23

2 Answers 2

0
$\begingroup$

Let us ignore your statement plates are considered as infinite.

You certainly have highlighted an apparent paradox.

  • Taking a positive charge from plate $P1$ to plate $P2$ via the region between the plate $P$ requires positive external work to be done so the potential of plate $P2$ is greater than the potential of plate $P1$.
  • Taking a positive charge from plate $P1$ to plate $P2$ but not passing through the region between the plate $P$ requires no external work to be done as there is no electric field in that region, so the potential of plate $P2$ is equal to the potential of plate $P1$, ie zero.

The mistake is that there is an electric field outside the region between the two plates and so positive external work does have to be done when moving positive charge outside of region $P$ and it is the same amount of external as when moving charge in region $P$.

Now what about the infinite plate idea?
All paths require the positive charge to travel through region $P$ with positive external work being done.

$\endgroup$
2
  • $\begingroup$ But what if we take a test charge from infinity near plate 1 to it then there will be no electric field there and thus it's potential will be 0.......and if we do the same thing for the right plate too it's potential will also be 0 $\endgroup$ Jun 15, 2023 at 17:47
  • $\begingroup$ Talking about infinity is potentially very dangerous and often produces inconsistent/nonsense results. Might it be that the parallel plates split space such that the two parts have differing potentials? $\endgroup$
    – Farcher
    Jun 15, 2023 at 22:05
0
$\begingroup$

The pd between the plates is $\Delta V=\tfrac 1Cq\ $ in which $\ C=\frac{\epsilon_0 A}d\ $ with the usual notation.

This is the work done per unit charge taking a test charge across the gap between the plates, as the magnitude of the electric field there is $q/\epsilon_0$.

But the electric field due to stationary charges is a conservative field, so we ought to find the same potential difference if we take our test charge from one plate to the other by an 'external' path that goes 'round the outside', not across the gap. Your paradox, I assume, is based on there being no field outside plates of infinite extent.

There are at least two ways of resolving the paradox. One – which you won't like – is to point out that if the plates are infinite you'll never be able to get your test charge beyond their edges because their edges are infinitely far away. Are we even allowed to say anything about the external field beyond the edges of the plates?

A less controversial way is to consider plates that, though much greater in length and width (or diameter) than the gap between them, are finite in size. With equally and oppositely charged plates the field is almost uniform in the gap except near their edges. The electric field lines 'bulge out' into the space beyond the plate edges, and if you take a test charge via an 'external' path from plate to plate you'll get the same value for $\Delta V$ as if you'd taken it straight across the gap in the central region (though direct calculation isn't easy because the 'external' field is non-uniform). It's only near the centre of the plates, that is well away from the edges, that the external field is zero. But to get from one plate to another via an external path, part of your path must go through a region where the field isn't negligible. Incidentally some charge will be found on the outside faces of the plates, near their edges.

$\endgroup$
8
  • $\begingroup$ But what if we take a test charge from infinity near plate 1 to it then there will be no electric field there and thus it's potential will be 0.......and if we do the same thing for the right plate too it's potential will also be 0 $\endgroup$ Jun 15, 2023 at 17:47
  • $\begingroup$ Interesting. We have two infinities: the dimensions of the plates and the infinite distance away. The problem is that if you're an infinite distance away (presumably on a line through the centre of the plates and normal to the plates), the plates won't any longer be infinite from your point of view and the field won't be exactly zero. In analysis of this set-up, use of the words infinite and infinity is, I think, best avoided unless their exact meaning in the context is carefully explained. – $\endgroup$ Jun 15, 2023 at 18:19
  • $\begingroup$ For two equal and opposite charged parallel discs I've just calculated the work per unit charge to bring a test charge from infinity (by which I mean a distance many times greater than the disc's diameter) to the centre of each disc along a normal to the disc. For the positive disc it is $\frac {Qd}{2A\epsilon_0}$; for the negative disc it is $-\frac {Qd}{2A\epsilon_0}$. This assumed that the separation, $d$ between the discs was much smaller than their diameter. The difference between the two potentials is, you'll note, the pd between the discs as calculated from a direct transit of the gap! $\endgroup$ Jun 15, 2023 at 22:17
  • $\begingroup$ Yes I got your point but when we have earthed the second plate should'nt potential there be 0......Btw initially it was not charged and only plate 1 is charge. So how will the charge distribution be? $\endgroup$ Jun 16, 2023 at 2:33
  • $\begingroup$ An equal and opposite charge to that on the charged plate will be induced on the facing surface of the earthed plate. In your diagram the induced charge is in the form of electrons that have flowed from the earth through the wire. In terms of potential my answers are based on two charged plates neither of which is earthed. Sorry. $\endgroup$ Jun 16, 2023 at 7:07

Your Answer

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

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