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If I apply Coulomb's Law, I get that the field is defined in all space for volumetric densities. For linear and surface charges, the field is undefined where the charges are (division by $0$).

Why wouldn't $ \mathbf{E}=\vec{0} $ inside an infinite line, or plane, as intuition suggest?

Why wouldn't $ \mathbf{E} = \vec{0} $ or $\mathbf{E} = \frac{\sigma}{\epsilon_0} \hat{n}$ in a closed charged surface (like a conductor)?

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    $\begingroup$ What do you mean by "in the charges"? $\endgroup$ – Joe Iddon May 13 at 15:02
  • $\begingroup$ If the charge is distrubuited in a line along the z-axis, "in the charges" would be at $z=0$. $\endgroup$ – MarcoCiafa May 13 at 16:07
  • $\begingroup$ I'm really sorry, my question was terribly redacted, I didn't even meant what I asked. Editing it. $\endgroup$ – MarcoCiafa May 13 at 20:06
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For linear and surface charges, the field is undefined where the charges are (division by $0$).

This is so only for linear charge distribution. For surface charge distribution, electric field can be defined everywhere (unless the distribution on the surface itself is divergent).

Why wouldn't $ \mathbf{E}=\vec{0} $ inside an infinite line, or plane, as intuition suggest?

Intuition is subjective. My intuition does not tell me that field is zero inside the line or a plane.

For charge on a line, it can be shown, using Gauss' law, that electric field diverges in the line; the closer to the line, the greater the field magnitude, increasing without any bound, while direction depends on the angle of approach. So there is no limiting process by which we could find unique value in the line.

For charge on a smooth surface, there is a discontinuity in the field when crossing the surface, but the field can be defined even in the discontinuity. In case the surface is closed and conductive, in equilibrium the field inside vanishes and field in the surface is half the field just outside the surface.

This "completion of field definition" in the discontinuity is possible and meaningful because field is force and there is measurable finite force acting on the surface due to its charged state.

Why wouldn't $ \mathbf{E} = \vec{0} $ or $\mathbf{E} = \frac{\sigma}{\epsilon_0} \hat{n}$ in a closed charged surface (like a conductor)?

The correct formula is

$$ \mathbf{E} = \frac{\sigma}{2\epsilon_0}\hat{\mathbf{n}} $$ i.e. half the field at a point just outside the surface. This force acts on the surface as a kind of pressure from inside, as the charges push each other further away.

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  • $\begingroup$ Thank you! This finite and mesurable force, just aplies in the context of a smooth closed conductive surface (i.e. conductors)? Or could I use that same argument for whatever charged surface and define the field on the surface to be the average of the field on one side and the other (I've already "proved" their diference is always $\frac{\sigma}{\epsilon_0}\hat{n}$)? $\endgroup$ – MarcoCiafa May 14 at 1:40
  • $\begingroup$ When I said intuition, I probably was suggesting the bilateral symetry arguments. Which in my understanding, say the field must be $ 0 $ in a infinite homogeneous line or plane $\endgroup$ – MarcoCiafa May 14 at 1:43
  • $\begingroup$ The force will be finite even for nonconductors, if the charge distribution is "nice enough", i.e. no charge density singularities on the surface (some can be inside, such as point charge inside, that is fine). The rule of 1/2 applies only if field due to the charge distribution on the surface vanishes inside. This requires a special distribution of charge which happens for conductors in equilibrium, but it is not true in general. In general field inside is non-zero too and there is no relation between electric field just outside and surface charge density on the closest point of the surface. $\endgroup$ – Ján Lalinský May 14 at 11:03
  • $\begingroup$ For infinite objects, force on an element of the object cannot be determined, the Coulomb integral is not defined. Finite objects are fine, but the symmetry is lost - the farther from the body center, the greater the force. For a closed charge surface, there is no such symmetry, the force will point outwards. $\endgroup$ – Ján Lalinský May 14 at 11:07
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There is no "inside" for a plane or line, they are as thick as a point, therefore zero width. Secondly, infinite distributions would demand infinite energy to put together, they serve as idealizations for other problems or situations but intuition is not to be followed (no way of imagining real infinite quantities). Thirdly, Gauss law would fail if we take a surface around a line distribution (a cylinder) and assume that for $\rho \rightarrow 0$ the field is null. Instead we calculate $\vec{E}$ for a given radius y then see what happens if that radius is small. As in the case of a point particle we have divergences where the charges are located. When one moves to calculate volumetric charge distributions we take the field to be made of a continuum of charges, avoiding the point particle divergence. This is justified in many ways, one is to calculate $\vec{E}$ at a point as if no charge was there (if there were the force produced by other particles would move it), also the continuum is produced when we change sums to integrals since the $\Delta q$ produced by each charge is just too small.

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