# Why is the electric field of an infinite insulated plane of charge perpendicular to the plane?

I'm studying Gauss' Law, and I came across a section where we're supposed to find the electric field of various shapes (like an infinite line of charges, etc), and for an infinite plane with a uniform positive charge per area, it says here in my notes:

Planar symmetry => Charge distribution doesn't change if we slide it in any direction parallel to the sheet => At each point, the field is perpendicular to the sheet, and it must have the same magnitude at any given distance on either side of the sheet.

It's not clear to me why having a charge distribution that doesn't change will result in a field perpendicular to the sheet. Can anyone help me clarify?

• If the charge distribution is uniform, what other direction than perpendicular coulda uniform field take? Commented Jul 1, 2014 at 14:55
• I think he would say there could be a component in the plane. Commented Jul 1, 2014 at 14:55
• Related: physics.stackexchange.com/q/441005/2451 and links therein. Commented Nov 15, 2018 at 9:29

The answer by @BrianMoths is correct. It's worth learning the language used therein to help with your future studies. But as a primer, here's a simplified explanation.

Start with your charge distribution and a "guess" for the direction of the electric field.

As you can see, I made the guess have a component upward. We'll see shortly why this leads to a contradiction.

Now do a "symmetry operation," which is a fancy phrase for "do something that leaves something else unchanged. In this case, I'm going to reflect everything about a horizontal line. I mean everything.

The "top" of the sheet became the "bottom." This is just arbitrary labeling so you can tell I flipped the charge distribution. The electric field is flipped too. (Imagine looking at everything in a mirror, and you'll realize why things are flipped the way they are.)

Hopefully, everything is okay so far. But now compare the original situation with the new inverted one.

You have exactly the same charge distribution. You can't tell that I flipped it, except for my arbitrary labeling. But if you have the same charge distribution, you ought to also have the same electric field. As you can see, this is not the case, which means I made a mistake somewhere.

The only direction for the electric field that does not lead to this contradiction is perpendicular to the sheet of charge.

You have to realize that the system is invariant under rotations about the normal to the plane. Then electric field must also be invariant under these rotations. An electric field component in the plane does change under such a rotation, so such a component must not exist if we have this invariance. Thus the electric field is purely along the normal to the surface.

Pick a point above the plane.

From a point in the plane directly under the point above, draw a circle of some radius.

Consider the contribution of the charge elements along the circle to the electric field at the point above the plane.

Since the charge density is uniform, the horizontal components of the electric field from charge elements on opposite sides of the ring cancel leaving only the vertical components which add.

Since this holds for all such rings, the electric field at the point above the plane must be perpendicular to the plane of charge.

An answer connected to Gauss law (I hope everything is correct, since it's long ago for me ... so no warranty):

An infinite plane of uniform charge for example in the z-plane has the charge distribution:

$\rho=q\,\delta(z)$

Thus, the electrostatic potential should be $\Phi=\frac{q\,|z|}{2\pi}$. Hence, the electric vectorfield is: $\vec{E}=\text{grad}(\Phi)=\{0,0,\frac{q\,\text{sign}(z)}{2\pi}\}$

In mathematica I can plot this like:

Show[ContourPlot3D[z == 0, {x, -1, 1}, {y, -1, 1}, {z, -0.5, 0.5}], VectorPlot3D[{0, 0, z/(2 \[Pi] (z^2)^(1/2))}, {x, -1, 1}, {y, -1, 1}, {z, -0.5, 0.5}]]


This can of course be generalized. Hope this helps you.

• The problem is essentially one-dimensional, so you should be using the 1-D Green's function for the Poisson equation, proportional to $|z|$, not the 3D one proportional to $|z|^{-1}$. The potential is linear, and the field strength constant (with a jump at the plate changing its direction). Commented Jul 2, 2014 at 10:11
• You are totally right, I thought about this yesterday too, but I was too lazy to solve it ... I'll changed the answer accordingly.
– PeMa
Commented Jul 2, 2014 at 12:20
• ... hope all signs are correct.
– PeMa
Commented Jul 2, 2014 at 12:27

With respect to your test charge, there will always be an equal number of charges on your plane in all directions because the plane is infinite. So for every charge "in front" of your test charge there will be a charge "behind" your test charge. And for every charge to the left of your test charge will be a charge to the right. What this means is that no matter where you place your test charge, the parallel force between the test charge and any particular charge on the plane is going to be cancelled out by the parallel force between the test charge and some other particular charge on the plane. That's how I intuitively think about it, at least.

Assume a component in the plane of the feild
Now we have to contradict the existence of this component
Then take a closed path in form of a rectangle which has two of its sides perpendicular to the plane and the sides which are parallel to the plane are in the direction of assumed direction of parallel component of feild.
Now work done in this closed loop is not zero.
Hence there is no component parallel to the plane.