# Shileding of a point charge by an infinite conducting plane

## Main Question or Discussion Point

Hi everyone,

I have some queries regarding the situation where a point charge is placed at a distance from an infinite conducting plane:say the plane is the xy plane and the charge is above it at distance z:

1) If the plane is at non-zero potential, is that true that the plane will not shield the electric field generated by the point charge, this is to say that electric field will be non zero for z<0? if so, is this due to the induced charges on the side opposite that facing the charge?

2) if the plane is instead grounded, the point charge is shielded (E=0 for z<0): is this because of the induced charge being discharged on the ground and hence no sources are there for generating field?
Thanks a lot!

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Good questions.
2) if the plane is grounded then the charge is shielded below the plate. Absolutely. (Use the method of images if you want to find the fields and potentials resulting).

I felt pretty good about the answer to 1)... but now i'm doubting, and confused on the details. I'm sure the conclusion is correct - the charge will not be shielded - but i'm not entirely sure why. I don't think its because of the charge on the opposite face (in and of itself at least), because the charge buildup is what does the shielding in the grounded case.
Also, because its a plane, i don't think you can have different charges on different sides (because its inifinitely thin), again i'm not positive about that.
Hopefully someone else knows better, now i'm really curious!

I think that the conductive plane shields the charge's field on the opposite side in any case: the induced charges in the plane will move until the field component parallel to the plane is zero: so the potential will be constant (C) on the whole plane. If C=0 then the potential on the opposite side must also be 0 (to satisfy Poisson(Laplace) equation and boundary conditions).

If C is nonzero, then the field in the whole space can be calculated as a sum of the field at C=0 and the field of the (uniformly charged) plane in empty space (because the Poisson equation is linear). However the field of an uniformly charged infinite plane in empty space has infinite range, so it is impossible to charge it with any finite potential (if potential is measured relative to infinity).

ok, thanks both. although, I must say I am not sure about the case when the infinite plane is not grounded, at a finite potential that is. But I guess Lojzek ought to be right, his argument does make sense. Would it be the same if, instead of a plane, we had an infinite conducting slab of finite thickness (so the charges on the other face, that facing z<0, would definetely exist?
thanks

rbj
ok, thanks both. although, I must say I am not sure about the case when the infinite plane is not grounded, at a finite potential that is.
what would you ground the infinite plane to? if it's truly infinite and homogenously conductive throughout, i would say your infinite plane makes a better candidate for being the "ground" reference than anything else. i think Lojzek is correct.

I think that the conductive plane shields the charge's field on the opposite side in any case: the induced charges in the plane will move until the field component parallel to the plane is zero: so the potential will be constant (C) on the whole plane. If C=0 then the potential on the opposite side must also be 0 (to satisfy Poisson(Laplace) equation and boundary conditions).
so only a moving charge would be detected in the conductive plane then?
considering that the charge is acting locally, would we detect a potential difference along the length of the conducting plane then?
Both on the opposite side.

and secondly, couldnt we affect the charge on the infinite plane in a localised manner?

sorry but still I am not convinced about not being there a difference between:
1) point charge over an infinite conducting slab (itself of finite thickness) put at zero potential
2) point charge over an infinite conducting slab (itself of finite thickness) put at potential V

In 1) I would not have field lines emerging from the slab facing away from the charge because of its ground state, whereas in situation 2) I think we do have a field emerging from the slab facing away from the charge (i.e. in the second half of the space cut by the plane) because of the charges that are there to keep the potential gradient across the slab and opposing those induced by the point charge.

In other words, I cannot see why we would have shielding in situation 2).
Thanks
Lor

2) I think we do have a field emerging from the slab facing away from the charge (i.e. in the second half of the space cut by the plane) because of the charges that are there to keep the potential gradient across the slab and opposing those induced by the point charge.
Electrostatic potential beween points A and B is defined as -integral(E*dr) from A to B. Only diferences in electrostatic potential have physical meaning, additional constant is arbitrary. If you say that an object A is charged to a potential V, but you don't say against which point V is measured, then you didn't tell anything about the physical system. Also I don't know why you mentioned "potential gradient across the slab": potential gradient inside a conductor is zero (if there is no currents). Whether the conductor is a plane or a slab is not relevant: if a plane causes zero field on the opposite side, then any additional conductors added to that side would remain neutral.

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so only a moving charge would be detected in the conductive plane then?
considering that the charge is acting locally, would we detect a potential difference along the length of the conducting plane then?
Both on the opposite side.

and secondly, couldnt we affect the charge on the infinite plane in a localised manner?
An infinite conductive plane perfectly shields the field of static charges. If the external charge moves then some field might get through the conductor. E-M waves can travel through a thin conductor, but their amplitude decreases exponentialy.

I believe I know the answer. A grounded conductor means its voltage is zero. In this case we are taking the voltage at infinity to be zero. Any conductor has the same potential and since in this case the conductor is infinite then its voltage should be zero. In another words any infinite plane is grounded. If we looked at the other part of space separated by the conducting plane we can guess that a good solution would be V = 0 everywhere, since this satisfies the boundary conditions and the charge distributions ( zero in this region ). Which means that the Electric field is also zero. If we assume that we replaced this plane by a conducting slab then the argument will be the same. There is still something which may seem puzzling: If we calculated the electric charge over the top face ( assuming the point charge is in z>0 half space) we will discover that it is q, the same value of the point charge. If the conductor was neutral then it should stay neutral because it is isolated from anything else. So the -q should rest on the other sides. But we discovered that E = zero on the opposite face. Which means sigma ( surface charge density =0) and q on this surface is zero. So where does the charge go?.. The charge will then go to the other vertical sides at infinity. and since as we go to infinity the electric field will go to zero, so the surface charge density will go to zero however since the vertical sides are of infinite area we can still get a finite charge -q.

A conclusion which seems very reasonable is that charges on the plate will try to escape to the ground even we started with a charge plate, the charges will still go to infinity and this is the good thing about grounding.

I want to add something to what is above.

If we have a finite plate which is large compared to distance between the point charge and the plate then we can assume that this is an infinite plate. If this plate is connected to some battery and the other terminal of the battery is connected far away to infinity. Then the plate will have a certain voltage V wrt infinity ( ground ). Therefore we can still solve the problem but now we will not add just an extra charge as the image/ we have to add a charged plate which produces a constant electric field and by changing the the surface charge density and the distance of the plate we can adjust the voltage of the equipotenial surface to be V. For the lower half plane we then have 2 boundaries: an infinite plate with voltage V and the voltage everywhere at infinity is zero. Therefore one should expect a potential gradient ( Electric field ) in the lower half plane. Actually understanding what ground is was not apparent for me until I checked Jackson's book. Please refer to it and you will find a description of "ground" in the footnote in I think the second or third chapter ( Boundary value problems ).