Infinite grounded plane induced surface charge

  • #1

Main Question or Discussion Point

I have a question about calculating induced surface charge:

Consider a system composed of an infinite grounded conducting sheet located at z=0 and a point charge positioned some distance d above the sheet.

Via method of images, we can find the potential in the region z>0. When we are done, we will have V(x,y,z) for z>0.

Then, we typically proceed to find the induced surface charge on the conducting plane. We do so by noting that, immediately above the plane, the field only has a normal component, which is [tex] \sigma / \epsilon_0 \hat{n} [/tex].
Since

[tex]E= -\nabla V[/tex]

and E only has a component in the normal direction (i.e. the z direction)

[tex]E = -\frac{\partial V}{\partial z} \hat{n}[/tex]
and so

[tex]\sigma= -\epsilon_0 \frac{\partial V}{\partial z}[/tex]

where the partial derivative is evaluated at (or immediately above) z=0. We can easily evaluate this derivative since we know V(x,y,z) (from the method of images above).

However, this process of obtaining [tex]\sigma[/tex] assumes that the field directly below the conducting sheet is zero--this assumption is present in that we had to use the fact that the field directly above the plane is [tex] \sigma / \epsilon_0 \hat{n} [/tex] and we can only argue that that is true if E=0 below the sheet (the argument would go as follows: We know E always undergoes a discontinuity of [tex]\sigma/ \epsilon_0[/tex] across any surface boundary. Since E is zero below the infinite sheet, the entire discontinuity must be due to the field above the sheet, and hence the field above the sheet is [tex]\sigma / \epsilon_0[/tex]).

But how do we know that the field below the sheet is zero? If it is because the sheet shields the region below the plane from static fields, how do we know that it shield's the region?
 
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Answers and Replies

  • #2
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Good question , and i will try to answer it , Lets say we have a point charge q outside a solid spherical conductor , it will pull minus q and have +q on the other side. The charge in the sphere will move around to cancel the E field inside the conductor. Maybe this is similar to the infinite plane. I don't know if i answered your question, hopefully some one can give you a better answer . And i wonder if its similar to being in a Faraday cage on the other side of the plane because it is infinite.
 
  • #3
394
15
E is zero because the electric field inside a conductor under static circuntances always is. The reason is simillar to what "cragar" said but the infinite plane conductor is an idealized situation that may cause some awe though.
 
  • #4
Thanks for your responses. I don't think we have solved the problem yet, though. The region below the plane isn't inside of a conductor in this case, since the conducting sheet lies in the plane at z=0. The region z<0 is free space. So we can't argue that E is zero because it is inside of a conductor--unless being below an infinite sheet is in some strange way equivalent to being enclosed in a cavity surrounded by a conductor, which could make some sense.

I guess you could change the situation a little and give the sheet a little thickness. Then, in the region directly below the plane the field would be zero because as cragar and facenian said, the field inside of a conductor is zero. And then you could shrink the thickness of the sheet to zero to get the infinitly thin sheet. But, then you couldn't argue that the total charge induced on the sheet is -q (as is often done in this method of images problem) because you would also have to account for the charge induced on the bottom layer of the sheet (since the sheet now has a thickness). In the limit as the sheet thickness shrinks to zero, the two charge distributions--the one on the top layer and the one on the bottom layer of the sheet--would merge into one and you would have to account for both in order to figure out what the total charge on the infinitely thin sheet is.
 
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  • #5
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15
You are partially right. The problem is when you take the limiting case the charge on both sides shouldn't merge you must think that you still have a conducting media in between otherwise you wouldn't have an infinite "conductor". The induced charge on the other side (below) is iqual to to charge on the upper side(otherwise the field won't cancel inside) and you have an electric field below and outside de conducting plane witch is the specular image of field over the plane
 
  • #6
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I may not have cought all the points in the above posts because I just read through it very quickly, but doesn't this just mean that the solution for the charge density is ambiguous until you specify the field behind the conducting plane? Like having a PDE without enough boundary conditions to determine the solution uniqely?
 
  • #7
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15
but doesn't this just mean that the solution for the charge density is ambiguous until you specify the field behind the conducting plane?
No, saying you have a conducting plane is the same as saying that the field is equal zero immediatly below the charge surface, you don't need to specify the field.
I you need detemine the filed in the semi-space over the charge surface nothing more need be said, however what happens in the semi-space below the charge surface depends on whether you have an infinite conducting media filling all the semi-space or just a plane conductor infinitely thin
 
  • #8
I thought I would tack this post on for future reference in case anyone who is interested in this topic reads this thread later:

There is also another way to see that the field in the region below the infinite grounded conducting sheet is zero: you can solve Laplace's equation for the region below the plane. The boundary conditions would be:

1) V=0 at z=0 (since the conducting plane at z=0 is grounded)
2) V=0 at infinity

As long as we are assuming that there is no charge below the plane, then that entire region satisfies LaPlace's equation:

[tex] \nabla^2 V = 0[/tex].

The solution to this equation with these boundary conditions is V=0 everywhere (in the region below the plane). Since V is constant, E is zero.
 
  • #9
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The equation
[tex]\nabla^2 V = 0[/tex]
does not say that there is no electric field in a region. Remember that a point charge satisfies that same equation everywhere except for at the point charge but the electric field is pervasive. If one were to construct an infinite, neutral, conducting plane and brought in a +q charge from infinity, by Gauss' Law, any surface integral of the electric field that encapsulates the entirety of the plane and the point charge must give +q/e_0. So, in fact, there will be an electric field below the conducting plane.

The issue with this problem is that one generally does not worry about the field below the conductor because we want to know only how bringing a charge near an infinite plane induces a charge on one surface. Solutions to PDEs like Poisson's Eq. above require only that you have a specified subset of space and you have boundary conditions specified on that subspace. In the case of the infinite conducting plane, our subspace is z>=0 and the boundary conditions are E = 0 at z=infinity and E=0 at z=0- where 0- means approaching 0 from below. Since the rest of space is not closed by our boundary surface, it is not necessary to consider it. The field for z<0 requires knowing more about the conductor.
 
  • #10
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15
The equation
[tex]\nabla^2 V = 0[/tex]
does not say that there is no electric field in a region. Remember that a point charge satisfies that same equation everywhere except for at the point charge but the electric field is pervasive. If one were to construct an infinite, neutral, conducting plane and brought in a +q charge from infinity, by Gauss' Law, any surface integral of the electric field that encapsulates the entirety of the plane and the point charge must give +q/e_0. So, in fact, there will be an electric field below the conducting plane.
No, there is no field in the conductor by definition. One formal way to see it is to solve Laplace's equation for the region inside the conductor like AcidRainLiTE did.
To determine the field for z<0 you said that more is needed to know about the conductor, that may be so if you want to know the field for the entire semi-space z<0 but in any case the field inside the condutor will be zero.
 
  • #11
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If you reread the quote you took from my response, you will see that I never contested that E=0 in a perfect conductor; I said below. The *more* that needs to be known about the conductor is how thick it is because outside this thickness, there can certainly be a field.
 
  • #12
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If you reread the quote you took from my response, you will see that I never contested that E=0 in a perfect conductor; I said below. The *more* that needs to be known about the conductor is how thick it is because outside this thickness, there can certainly be a field.
ok, we all agree now.
 
  • #13
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Well I do not agree...

@ednobj states that Laplace's equation holds in a region that encloses a point charge, which is obviously not true...

Moreover he's reasoning with Gauss's law a false assumption, namely that there would be a field below the plane. The problem is, I cannot tell why this is false!

This topic brings up a question not yet answered! WHY the electric field of a point charge "vanishes" below an infinite plane conductor, regardless of its thinness?
After all, the range of the electromagnetic force should be infinite...
 
  • #14
haruspex
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WHY the electric field of a point charge "vanishes" below an infinite plane conductor, regardless of its thinness?
After all, the range of the electromagnetic force should be infinite...
It vanishes for the same reason it does inside a conducting shell. It is exactly canceled by the fields from the induced charges above it.
 
  • #15
Born2bwire
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Well I do not agree...

@ednobj states that Laplace's equation holds in a region that encloses a point charge, which is obviously not true...

Moreover he's reasoning with Gauss's law a false assumption, namely that there would be a field below the plane. The problem is, I cannot tell why this is false!

This topic brings up a question not yet answered! WHY the electric field of a point charge "vanishes" below an infinite plane conductor, regardless of its thinness?
After all, the range of the electromagnetic force should be infinite...
There is a field below the conductor, they are not disagreeing with that point. Laplace's Equation does hold in a region where there are point charges, except at the locations of the point charges. That's because the charge distribution is one of dirac-delta's and so the right-hand side of the Poisson Equation is only non-zero at the exact location of the charges. ednobj makes this distinction in his post as well.
 
  • #16
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@Born2bwire there is no field below the conductor and the reason why is the fact that the field of the conductor produced by electrostatic induction compensates the field of the point charge.

Moreover, we are talking about regions of space here. There is no point in saying that Laplace Equation is applicable to this case, cause the region we are interested in, contains a charge!

His Gauss's law argument is probably false because he doesn't take the field of the conductor into account.
 
  • #17
Born2bwire
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@Born2bwire there is no field below the conductor and the reason why is the fact that the field of the conductor produced by electrostatic induction compensates the field of the point charge.

Moreover, we are talking about regions of space here. There is no point in saying that Laplace Equation is applicable to this case, cause the region we are interested in, contains a charge!

His Gauss's law argument is probably false because he doesn't take the field of the conductor into account.
It seems that there has been a bit of confusion throughout this discussion. It is true, now that I look at the conditions given in the OP that there is no field below the conductor. However, the reason you state is not the whole story. Take the case of a thin conducting spherical shell surrounding a point charge. The field outside of the shell is the same field as if there was no shell to being with. There is no field INSIDE the conductor, but there is a field outside the conductor. This is because of charge conservation. The source charges induce a surface charge on one side of the conductor (let's say the inner side). This requires that charges from the outer side of the conductor to migrate toward the inner side. This leaves another induced surface charge on the outer surface. So the induced charge does cause external fields, which we can confirm via Gauss' Law.

There is one catch here that was missed though. The problem assumes a grounded conductor. If the conductor is grounded, then there is no field below the sheet. That is because we no longer have charge neutrality on the conducting sheet. As a surface charge is induced on the top edge of the sheet, the induced charge is pulled from the ground reservoir and not the opposite side of the conductor. In essence, the sheet now has a net charge. If the conductor is ungrounded, then there is a field due to the reasons given above. It appears that ednobj's discussion was not taking into account the grounding of the conductor.

Laplace's Equation works here because when dealing with the differential case we can solve the problem purely from the boundary conditions via the Uniqueness Theorem. The Uniqueness Theorem states that we we can uniquely define the potential inside a region by specifying the charge density inside the region and the value of the potential on the boundary. When they are talking about finding the field below the conductor, we are talking about a region that is not enclosing a charge, so Laplace's Equation is perfectly valid here. If we were talking about an ungrounded conductor, then we could encapusulate the effects of the charge above the sheet by specifying the potential along the sheet. Even in the case of enclosing a region with a charge, we still talk about solving Laplace's Equation because we often are looking at solving along a boundary where there is no charge. In the case of finding the potential above the sheet, we set the boundary conditions to be zero on the surface of the sheet and at infinity. The problem then becomes one of solving for the Laplace's Equation along this boundary. The simple solution, inspired by image theory, is to use the potential due to a point charge which you will note satisfies the Laplace Equation at locations outside of the charge. The placement of the charge and its image is dictated by our knowledge of the charge distribution above the plate.
 
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  • #18
haruspex
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I doubt it matters whether the plate is grounded. There is no field below the plate in either case.

The potential is the same everywhere in the plate. Any field below the plate must be a result of a different potential somewhere below it. If we posit a different (but finite) potential infinitely far below it then, yes, there's a potential difference. But since the distance is infinite the field is zero.
 
  • #19
Born2bwire
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I doubt it matters whether the plate is grounded. There is no field below the plate in either case.

The potential is the same everywhere in the plate. Any field below the plate must be a result of a different potential somewhere below it. If we posit a different (but finite) potential infinitely far below it then, yes, there's a potential difference. But since the distance is infinite the field is zero.
It matters from the perspective of charge neutrality. If the plate is ungrounded and uncharged, then the net charge must remain zero. Placing a charge above the plate induces a surface charge on the upper side of the plate to satisfy the boundary conditions. If that occurs, then there must be another region on the plate that has the opposite charge distribution to allow the net charge on the plate to remain zero. This would occur on the underside of the plate.

We can easily take the case of an infinite sheet of charge above an ungrounded infinite sheet of conductor. We know that a sheet charge is induced on the upper half of the PEC sheet. So if the sheet of charge has the charge density of σ, then the near side of the PEC sheet gets -σ/2 induced and the far sheet gets σ/2. In this way the PEC sheet still has no net charge but we see that the fields become:

Region 1


---------------------------------------------- σ


Region 2


========================== -σ/2
Region 3 - PEC
========================== σ/2

Region 4



The fields become:
[tex] \mathbf{E} = \left\{ \begin{matrix} \frac{\sigma}{2\epsilon_0} \hat{z} & \qquad \mbox{Region 1} \\ -\frac{\sigma}{2\epsilon_0} \hat{z} & \qquad \mbox{Region 2} \\ 0 & \qquad \mbox{Region 3} \\ -\frac{\sigma}{2\epsilon_0} \hat{z} & \qquad \mbox{Region 4} \end{matrix} \right. [/tex]

We can use Gauss' Law and see that it is properly satisfied for this solution.

Now if we were to ground the PEC sheet, then it would always have a voltage of zero. Below the sheet all the boundary conditions for the region would have the potential set to zero and since there would not be any free charge below the sheet we know that only the trivial solution to Laplace's Equation exists, that the potential and fields are all uniformly zero below the sheet. Physically what happens is that the ground reservoir will source electrons to the PEC sheet so that the near side of the PEC sheet will have a charge distribution of -σ.
 
  • #20
haruspex
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The original post concerned fields induced by a point charge, not an infinite sheet of charge. According to my calculations, an infinite sheet with a nonzero uniform charge density would produce an infinite potential everywhere, so isn't a useful model.

With a point charge near an infinite ungrounded conducting sheet, you get some distribution of charge which results in a uniform potential everywhere in the sheet.
If you now ground it, what will happen? Would you agree a finite quantity of charge would flow to/from the sheet? Moreover, for the potential to remain constant across the sheet, the change in charge per unit area would be the same everywhere. Finite charge divided by infinite area = no change in charge density anywhere.
 

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