1
$\begingroup$

Setup 1: the grounded case

When a positive point charge is near a conducting grounded infinite plate of finite thickness, the area near the charge becomes negative.

The net charge on the plate should be negative:

One way to find the total charge on the plate is to use the method of images (replace the plate with an image, negative charge on the other side of the plate). The potential in space is just the dipole potential (it vanished on the plane of the right side of the plate). From the potential we can derive the electric field, and from it the charge density on the plate.

enter image description here

The potential inside the plate vanishes. The area left to the plate has no charges, and the boundary conditions are those of vanishing potential, so the solution is trivial: the potential vanishes left of the plate. This means that there is no electric field there, and so no charge density on the left side of the plate.

This means that in total we have negative net charge (which can be found by integration). The result is the same if the thickness is zero.

So far so good. But could this be more subtle?

Setup 2: the neutral case

A claim that the plate could be uncharged:

This article claims that given an uncharged, non-grouded, conducting infinite plate gives the same electric field as a dipole, using the method of images. If the fields are the same as in the grounded case discussed above, then the potentials in both setups must be the same (let the potential at infinity vanish). So the plate is de-facto grounded, even if not physically.

But that must mean that the charge densities on the plate are the same in both setups!

enter image description here

The article does agree that the charge density on the left side of the plate vanished, but it claims that the net charge might still be finite on it (and positive).

The claim is that as we take the limit of larger plate, the charge is constant (and positive), and the area grows, the charge density is uniform, and goes to zero.

How do we resolve this contradiction?

Which of these claims is false?

  1. Setup 1, the grounded case, has a negative net charge.
  2. The article claims that the uncharged, non-grounded setup 2 has a dipole field, and this means that the fields of the two setups are equivalent.
  3. This means that the charge densities are the same in both cases.
  4. This means that the net charge must be the same, which means that the article is wrong in assuming that an uncharged non-grounded plate near a point charge would have a dipole field.

I have some thoughts on how to solve this contradiction, but I'm not sure:

  1. The article is mistaken in the assumption that the plate has zero potential. A large, uncharged plate near a point charge must have some nonzero potential. Looking at an infinite plate is problematic if its potential is nonzero, because the boundary conditions are contradictory: the potential at infinity must equal that of the plate, as it extends to infinity.
  2. For an uncharged plate, the charge density on the left side is not uniform (unlike the article's claim). For a thin plate, the positive charges gather on the outskirts of the plate, far away from the positive point charge. the negative charge stays in the center. This means that we push finite charge off to infinity and that could create an ill-defined setup, which undermines some of the assumptions used to calculate the potential.
  3. The question of net charge is undefined in the infinite-plate setup, and more subtle than just integrating over the charge density. This means that either the first setup of a grounded plate actually has zero net charge, or that we just cannot find out the net charge (what a strange option!).
$\endgroup$
8
  • $\begingroup$ The net charge on the plate is not negative. It is zero. All that has happened is electrons moved toward the side of the plate near the external positive charge, leaving the other side of the plate positively charged. In order for the plate to have an overall negative charge positive charge would have to leave the plate. $\endgroup$
    – Bob D
    Commented May 14, 2023 at 17:45
  • $\begingroup$ @BobD... or negative charge moves from the ground to the plate. Isn't this how neutral objects are charged by electrostatic induction? I would also argue that in the real world the act of connecting a neutral plate to ground would give the plate a small negative charge by itself, but that's nitpicking. (To the OP: infinite and semiinfinite planes are always problematic) $\endgroup$
    – Peltio
    Commented May 14, 2023 at 21:51
  • $\begingroup$ My comment was on the first paragraph of your post and the bold statement that immediately followed, which didn’t mention grounding $\endgroup$
    – Bob D
    Commented May 14, 2023 at 22:32
  • $\begingroup$ @BobD In the second case, the one which the article in the link describes, the plate is uncharged and non-grounded. I've edited the post to make the contradiction more clear. $\endgroup$
    – Rd Basha
    Commented May 15, 2023 at 7:24
  • $\begingroup$ @Peltio Yes, infinite planes are sometimes problematic. I'm trying to pin down exactly what is wrong. I think that setup 1 is fine, and negative. I'm not sure about setup 2, what exactly happens when the plate is not grounded? My thoughts are in the post. $\endgroup$
    – Rd Basha
    Commented May 15, 2023 at 7:27

1 Answer 1

0
$\begingroup$

I've thought about it, and will answer my own question.

The solution is the third option: we cannot simply integrate the charge density, and conclude that this is the total charge.

One way to see this, is by looking at a charged, non-grounded, conducting sphere of radius $R$ and charge $Q$, at a distance $d$ from a charge $q$.

We can use two image charges inside the sphere, to get the potential.

$$V= \frac{Qd+qR+QR}{R(R+d)}$$

This potential goes to zero as $R\rightarrow\infty$, for all $Q$.

This means that an infinite plate with vanishing potential (e.g. a grounded plate) could have arbitrary charge. Integration of the charge density we get from the fields gives a total charge of $-q$. The remaining charge is far away, but adds nothing to the potential.

$\endgroup$

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.