0
$\begingroup$

My question is very similar to this one: Why is the field inside a hollow sphere zero?, but it's generalized to a non-spheric conductor, since we know that electrostatic field in the empty space (I'm not referring to the middle of the conductor, that's clear to me) of a hollow conductor is zero irrespective of the shape of the conductor. Could you show me why (even intuitively)?

EDIT: I try to be more specific. We have the 2 situations in the image (no charge inside). In case A) the geometric reason why the electrostatic field in the hollow is zero is described in the thread I quoted before, but that explanation is true only for spheric conductors. Nevertheless, we know that the field in the hollow is zero irrespective to the shape of the conductor so it has to be zero in case B) too. How do you explain that by vectorial sum? My intuitive hypothesis is that the irregular shape produces not uniform (according to different radii of curvature) charge distribution on the surface and this two effects (irregular shape and irregular distribution) balance each other in order to cancel internal field in every point as the sphere does. Is that true? image

$\endgroup$
5
  • 1
    $\begingroup$ It isn't. It's only zero if there are no charges in it. $\endgroup$ Commented Oct 6, 2022 at 22:46
  • $\begingroup$ Thinking about it in terms of charge makes it a complicated problem. Thinking about it in terms of potential makes it simple. $\endgroup$
    – John Doty
    Commented Oct 7, 2022 at 11:42
  • $\begingroup$ But electrostatic potential is just a consequence of the presence of charge, isn't it? $\endgroup$
    – max
    Commented Oct 7, 2022 at 14:40
  • $\begingroup$ @JohnDoty Anyway, I understand that, but my focus was on an intuitive explanation of how can the field be zero in a non-symmetrical situation in presence of charges (on the surface of the conductor) that should generate an electric field all around them, not only towards outside. $\endgroup$
    – max
    Commented Oct 7, 2022 at 14:41
  • $\begingroup$ Yes, but it's difficult, in general, to work out the charge distribution on the surface of a conductor, where the charges are free to move. The only way I know to do it is to back the distribution out from the field, which, in this case, you can get from the potential. $\endgroup$
    – John Doty
    Commented Oct 7, 2022 at 16:35

5 Answers 5

1
$\begingroup$

As FlatterMann says, you must assume there are no charges inside. You also must assume static conditions (no electromagnetic waves). The conductor is necessarily at constant potential, since if it isn't, it's not static: the potential difference will drive a flow of current through the conductor. There is thus no difference in potential between any part of the interior and any other part. With no potential difference, you have no field.

$\endgroup$
3
  • $\begingroup$ just to clarify, these exercises assume a perfect conductor is that correct? $\endgroup$
    – antimony
    Commented Oct 7, 2022 at 1:07
  • 1
    $\begingroup$ @antimony yes, I'm assuming that $\endgroup$
    – max
    Commented Oct 7, 2022 at 10:53
  • $\begingroup$ Can you elaborate on why this is: "There is thus no difference in potential between any part of the interior and any other part". The hollow part is not conductive, so your potential argument doesn't work $\endgroup$
    – Joe C.
    Commented May 7, 2023 at 21:41
0
$\begingroup$

Suppose first of all that the conductor were solid. Under electrostatic equilibrium, there will be no field in the interior of the conductor because if there were any field, it would make charge carriers inside the conductors move around until there is no net field.

From the fact that $E = 0$ in the interior of the conductor, we also deduce that $\rho = \nabla \cdot E = 0$ in the interior. That means that even if the conductor carries any net charge, that charge must all go to the surface. Similarly, if there is any static external field, it will be cancelled in the interior of the conductor solely by induced charge on the surface.

Now, just remove the material from the interior. I don't mean reach in and remove it physically; I just mean that since this configuration with all charges on the surface satisfies Maxwell's equations, the same will be true of a hypothetical alternative configuration where the material in the interior of the conductor (which, recall, has zero charge density) is replaced by neutral air. Here the electric field will also be zero because the air has the same charge density as the material that it replaced (namely zero).

(As others have pointed out, if there are any charges in the interior of the hollow conductor, the field inside the conductor will not be zero.)

$\endgroup$
2
  • $\begingroup$ When you 'remove' the material of the conductor from the interior, I think you're changing initial assumptions (solid conductor) on which your statement ("Under electrostatic equilibrium, there will be no field in the interior of the conductor because if there were any field, it would make charge carriers inside the conductors move around until there is no net field") depends. Am I wrong? $\endgroup$
    – max
    Commented Oct 7, 2022 at 11:07
  • $\begingroup$ @max The scenario with the solid conductor is only used to derive some properties of a solution to the laws of electromagnetism. In that configuration, Maxwell's equations are satisfied and each charged particle feels no net force. This tells us that there exists an equivalent configuration where the conductor is hollow, since nothing about replacing the interior material with air causes the equations to stop holding. $\endgroup$
    – Brian Bi
    Commented Oct 7, 2022 at 12:13
0
$\begingroup$

I think gauss law may have a good explanation to you, this law states that flux(Electric field passing through unit area) is equal to total charge enclosed within surface over epsilon not where epsilon not is the permativity of free space,also now matter how much charge you give to the conductor the extra charge always remains upon the outer surface of conductor never inside, since the charge enclosed by inner surface is zero so is the flux, now if the flux is zero it could mean that maybe electric field and area vector(vector perpendicular to area) are perpendicular to each other or the electric field is zero. Now you can clear the area vector being perpendicular to electric field since this conductor is subjectedemphasized text to its own field not external

This figure shows area vector enter image description here

Now the cylendrical figure you see is placed within external electric field so you can see that field lines are perependicular to some of the surface of cylinder like curved surface, but in the sphere problem the surface is experiencing its own electric field which can be never perpendicular

$\endgroup$
0
$\begingroup$

You can think of metal as a material of $\infty$ dielectric constant.

Now, $\vec{E}_d=\vec{E}\left(1-\frac{1}{k}\right)$, where $E_d$ is the opposing field dielectric produces(using induced bound charges) in response to the external electric field $E$. Such that the overall field inside the dielectric$=\vec{E}-\vec{E}_d$.

For metals, $k=\infty$. $\Rightarrow \vec{E}_d=\vec{E}$, so field inside the bulk of the material = $\vec{E}-\vec{E}_d=\vec{0}$.

$\endgroup$
0
$\begingroup$

The potential is constant at every point on the internal surface $S$ of the metal: $\phi(r\in S)=\phi_0$. Since there are no charges in the cavity, the potential $\phi(r)$ satisfies Laplace's equation $$\Delta \phi(r)=0$$

The solution $\phi(r)=\phi_0$ solves the equation with boundary condition $\phi(r\in S)=\phi_0$. Since the solution of Laplace equation with closed boundary must be unique (How do I show that the Laplace equation has a unique solution under the Dirichlet closed-surface boundary condition?), is the only solution and $\vec{E}=-\nabla \phi=0$ everywhere inside.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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