4
$\begingroup$

Suppose you know that the electric field in distance $r > R$ from the center of a charged sphere with charge $Q$ and radius $R$ is given by:

$$ E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} $$

(Coulomb's law).

From this I want to derive the field of electric field of a charged conducting plate with given charge density $\sigma := \frac{Q}{A}$ with the following heuristic argument:

I start from the following three remarks:

  • Suppose $r \approx R$
  • The sphere is locally flat
  • The field of the sphere is locally approximately homogenous

Then this local approximately flat small part the sphere should be the same as just a conducting plate.

Since the total area of the sphere is $A = 4\pi R^2$, I can conclude:

$$ E = \frac{Q}{4\pi r^2} \frac{1}{\epsilon_0} = \frac{\sigma}{\epsilon_0} $$

should be the electric field of a conducting sphere (if you are close enough to it and not at the boundaries of the plate).

However the correct formula is:

$$ E = \frac{\sigma}{2\epsilon_0} $$

Or from another point of view: The formula which I derived is correct for two parallel conducting plates with equal but opposite charge densities.

So what is wrong with my argument that I get a factor $2$ wrong? Is there any way to modify this argument to get the correct result in a clear and convincing way?

Edit As Phoenix87 commented: The difference of our local situation to to that of a plate is, that on the "local plate" (piece of the sphere) the field is zero on one side. If you have just a plate, you have an equal but opposite field on both sides. So the situation seems to be locally that of one plate of a parallel plate capacitor, then the formula above would be correct. But I think this argument is not very clear. Any ideas to improve it?

$\endgroup$
  • 1
    $\begingroup$ since the sphere is closed, it behaves like a shield and there is no electric field inside. on the other hand, for a plate you would have electric field on both sides, hence the factor of 2. To visualise this more clearly, the sphere has a finite radius, and by superposition you get the contributes to the electric field even from the antipodes, and all of this sums up to make up for that factor of 2. $\endgroup$ – Phoenix87 Jan 22 '15 at 15:51
1
$\begingroup$

This really interesting, I have never seen this reproduced in text books on electrostatics. I feel like the problem may be due to the fact that you are multiplying by an area $A=4\pi R^2$ as this will count the surface of the sphere on the opposing side as well as the surface you are immediately adjacent to.

If you consider "flattening" the sphere out you would have effectively doubled the charge density on the surface as you would have two positive charged surfaces (i.e. opposing poles of the sphere).

I am not entirely sure but I think the correct thing to do would be consider a hemisphere and use that which would give the right result.

If you consider a hemisphere locally (close enough to ignore the edge effects and so that the surface of the sphere appears planar) then "flattening" the sphere leads to a plate of area $2\pi R^2$ which would give you the correct result.

$\endgroup$
  • $\begingroup$ Does the hemisphere has the same field strength as the sphere? Why? Yes then I would get the correct result, but if it has half of the field strength, both "halfs" would cancel out... $\endgroup$ – Julia Jan 22 '15 at 17:09
1
$\begingroup$

Nice question! As has been said in the comments, the problem is that as you get very close to the sphere, you have to take into account all the charge present. More precisely, the field inside is zero no matter what, as opposed to a plate, in which the field is the same on both sides. This explains the factor of $2$.

We might expect that taking the limit $R\to \infty$ could rid us of this problem, since the opposite side of the sphere will be very far away, and shouldn't contribute to the field. But do we hold $Q$ or $\sigma$ constant?. If we make the sphere bigger and bigger while $\sigma$ is held constant, then $Q$ goes as $\sigma R^2$, and so while the far side of the sphere gets farther away, its charge becomes bigger so that the field is constant (since it goes as $Q/r^2$). Clearly this is not the way to go, since obviously keeping $\sigma$ constant means that $E=\sigma/\epsilon_0$ is not going to change.

So let's try hold $Q$ constant and let $R\to\infty$. But now $\sigma \sim Q/R^2$, so $\sigma$ goes to zero. The field is zero on both sides of the sphere. In a way, we did manage to get the correct limit, since the field for a charged plate with $\sigma=0$ is zero too!

You shouldn't expect being very close to the sphere to be indistinguishable from a plate, since you can't ignore the contributions to the field from far away charges! You can only do that when the field becomes infinite, which is not the case here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.