# Point charge above infinite conducting plane

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1. Oct 8, 2014

### stroustroup

This question seems to come up often, but I cannot find a satisfying explanation.

There is a point charge +Q some distance above an infinite conducting plane. Supposedly, the electric field below the plane must be zero. I have trouble understanding why this is true.

The total charge on the plane must be zero, since the plane is a neutral conductor, and conservation of charge implies that any induced negative charge on the upper surface must be cancelled by an induced positive charge on the lower surface. Thus, if you look at a point P *outside* the plane, but *below* it, the fields from the positive and negative charges should not cancel each other. Indeed, the field must clearly be different in the configuration

+Q *

- - - - - - - - - - -
P *
+++++++++++++

than in

+Q *

- - - - - - - - - - -

+++++++++++++

P *

because the field from the positive charges has now switched directions (but not the field of the negative charges).
But since the field is zero inside the conductor, then it must be nonzero outside.

Moreover, if you use Gauss' law on a box containing P and going across the plate surface (but not containing Q), the net charge must be 0 (the plate has no net charge), so the electric flux must be 0 too. But the electric field is nonzero above the plane, so it must be nonzero below it as well.

Another thing which confuses me is the meaning of "grounded". The problem is supposedly different if the plane is grounded or ungrounded. But is there a distinction between "grounded conductor" and "ungrounded conductor"? Grounded supposedly means that the potential is 0. But I thought the actual value of the potential was irrelevant, so I could decide that the potential would be 117 V on the plate instead, without affecting the solution.

2. Oct 9, 2014

### ehild

Grounded means that the plate is connected to the ground (it is usually realized by sticking a metal rod deep into the soil, where the soil is wet usually) and connecting the "Earth" lead to it. The ground is considered at zero potential with respect to infinity - you can consider the Earth as a spherical capacitor, but it is so big that its potential does not change appreciably if you add a few coulomb charge to it.

If you have a grounded metal plate and a charge Q above it, the surface facing to he point charge becomes oppositely charged by -Q, and the like charges are pushed to the ground.
If you have a finite metal plate which is not grounded, the opposite side becomes charged, it has the same Q charge. See picture,
On the upper face, negative charges accumulate so as the surface charge density is greatest just below the point charge and decreases farther away.
On the other side of the plate, the charge does not "feel " what is on the upper side. The charge distribution is homogeneous. The surface charge density is $\sigma=Q /A$ where A is the area. If the area is very big, the surface charge density is very small. At the limit of infinite area the surface charge density is zero. The electric field is $E=\sigma/\epsilon_0$ below the plate, zero at the limit of of infinite extension.

ehild

3. Oct 9, 2014

### stroustroup

I don't see why this should be true. Is it because the electric field inside the conductor is zero? But why wouldn't the positive charges be distributed symmetrically to the negative charges?

4. Oct 9, 2014

### ehild

The positive charges do not feel the negative charges on the upper side. What would make them distribute symmetrically? The electric field is zero inside the conductor plate. The surface charges on the bottom face feel only the repulsion forces from each other. So they distribute as far of each other as possible.

ehild

5. Oct 9, 2014

### stroustroup

Oh I see. Because of the point charge disturbing the conductor, the distribution of charges has to be non-symmetric if we want the electric field in the conductor to be zero, that makes sense.

So this also explains why the field is zero below the plane, since a finite charge of +Q is spread out infinitely far, and thus the positive surface charge contributes a zero electric field in the limit. It is the nonuniform negative surface charge distribution on the upper surface which does the job of canceling the field from the point charge Q.

So the result for the electric fields is the same whether or not the plane is grounded.

6. Oct 9, 2014

### ehild

True in case of infinitely big plate.
You can also say that the edge of the plate extending to infinity is at the potential of infinity, which is zero. As every point of a conductor is at the same potential, the whole plate is at zero potential. As there is no potential difference between it and the infinity, the electric field is zero between infinity and the plate on that side where no external charge is.