Is a Grounded Plane Electrically Neutral?

[Razvan]

Consider the following configuration: a grounded plane (very thin and with an infinite surface) and a positive (not important) point charge above it.

What does grounded really mean? Can't I assume that it is neutral and it has infinitely many positive and negative charges?

Because of the positive charge, electrons are brought on the top surface of the plane. But now is the plane neutral or "it received more electrons from the ground"? Because if it were neutral, the electric field above the plane would only depend on the point charge.

Thank you.

 

[vanhees71]

http://en.wikipedia.org/wiki/Method_of_image_charges

 

[Razvan]

That is where my problem comes from. So the charges that have accumulated on the top surface of the plane "come from the ground" and make the plane (if we consider the plane and the "ground" to be separate objects) electrically negative, generating an electric field outside.

If we considered the plane to be just a neutral object with infinite amount of charge, even though the charges would rearrange because of the point charge, it would not generate any electric field. Is this correct?

 

[vanhees71]

It would also generate an electric field because of the rearrangement. Also a non-conducting di-electric creates a field due to the induced dipole (and higher multipole) moment(s). You can solve the analogous problem for a dielectric plane (or a bit simpler a dielectric-filled half-space) with help of the image-charge method. You only have to adapt the boundary conditions to the new problem.

 

[Razvan]

But if Gauss' method is applied on a surface that surrounds the plane, wouldn't the charges inside cancel out?

 

[vanhees71]

Yes, but you have polarization. Googling gave the following nice review on Maxwell equations and boundary conditions:

http://local.eleceng.uct.ac.za/courses/EEE3055F/lecture_notes/2011_old/eee3055f_Ch4_2up.pdf

 

[Razvan]

So in order to get the correct electric field, the gaussian surface should be for example a very thin cylinder with one base inside the plane and the other outside?

 

[zoki85]

So in order to get the correct electric field, the gaussian surface should be for example a very thin cylinder with one base inside the plane and the other outside?

What cylinder are you talking about? In your example have a point charge q at distance d above conducting flat ground plane. In order to calculate electric field in space between plane and point charge q, put an oposite charge -q at distance 2d from point charge q ( symmetrially with respect to the plane surface) and pretend like the plane is not there.

 

[Razvan]

I was thinking if the plane had a very small thickness (but not flat), what would be the correct way of finding the electric field caused by the induced charges without using the image method. So I thought that a small cylinder (or rectangular parallelepiped) with one base inside the conductor and the other above would work. Is this wrong?

 

[zoki85]

I remind you said grounded plane:

Consider the following configuration: a grounded plane (very thin and with an infinite surface) and a positive (not important) point charge above it.

This makes the same field configuration (between point charge q and thin plane) as situation with charge q and conducting ground plane (of infinite thickness)

 

[Razvan]

If we considered the plane to be just a neutral object with infinite amount of charge, even though the charges would rearrange because of the point charge, it would not generate any electric field. Is this correct?

This was one of my misunderstandings: I thought that if the plane wasn't grounded, it wouldn't generate any electric field because it would be electrically neutral (which is wrong).

But making the plane grounded, and considering the outside charge q, we can say that the plane isn't neutral anymore, correct?

Saying the plane is grounded only means that its potential is the same as the potential of one point at infinity (0 V).

Now if the plane wasn't grounded, this boundary condition wouldn't be true and the electric field would be different. Is this correct?