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If you have a potential difference between two plates (a capacitor) then you have and electric field between the plates.

Is there also a perpendicular magnetic component to this electric field? I've never heard of this - but I thought electric and magnetic fields must always coexist in phase and orthogonal for every packet of radiation?

If the electric field is not electromagnetic radiation then how does the other plate "know" about the potential from the top plate?

Mark Corrigan
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3 Answers3

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Magnetic fields are created by changes in the Electric Field. If the Voltage difference between the plates is constant, so is the electric field, so there will be no induced magnetic field.

To answer your second question, the bottom plate doesn't need to "know" about the top plate. The potential difference is a consequence of the charge distribution between the two plates, taken together as the system.

Radiation is a self-propagating electromagnetic field - an oscillating in electric field that produces an oscillating magnetic field, that produces an oscillating electric field, ....

D. W.
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  • So if an AC signal is present this will create a magnetic field component orthogonal to the E field and also it will create propogating radiation at right angles to both of these components? – Mark Corrigan Sep 07 '16 at 20:38
  • Mark, that is right and this is how radio waves are created, see http://physics.stackexchange.com/questions/278657/visualizing-a-light-wave/278817#278817 – HolgerFiedler Sep 07 '16 at 20:45
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Electric field gives a birth to magnetic field only when it changes in time. This is expressed in Maxwell's equations:

$\nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}$

$\nabla \times \vec{B} = \frac{\partial \vec{E}}{\partial t} + const \cdot \vec{J}$

Static electric (magnetic) fields do not radiate.

Static electric field gives a birth to Coulomb force:

$\vec{F} = const \cdot \frac{q_1 q_2}{r^2} = q \cdot \vec{E}$

That is how the plates "know" about each other.

You confusion probably comes from a fact that EM-waves can be described by particles - photons, wheres in undergraduate physics course there is no word about such thing for static fields. Well, if you want to think about that this way, there are particles which are analogues of photons but for stationary fields - virtual photons.

MsTais
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Is an electric field electromagnetic radiation?

Electromagnetic radiation is a propagating disturbance in the electromagnetic field. This disturbance has an associated energy and momentum and so, electromagnetic waves transport energy and momentum through space.

So no, an electric field is not electromagnetic radiation.

However, static electric and magnetic fields have an energy density and crossed static electric and magnetic fields have momentum. The key is that this energy and momentum is not in transit when the fields are static.

If the electric field is not electromagnetic radiation then how does the other plate "know" about the potential from the top plate?

The electric field is a physical entity. An electrically charged particle here 'feels' (couples to) the electric field here now without 'regard' for the charges elsewhere that produced the electric field.

Alfred Centauri
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  • Just curious. As per your definition of radiation, if i have a uniform electric field in all space, and i suddenly switch its magnitude 10 times, then this will create a 'disturbance' propagating throughout the field, correct the previous value there, to 10 times its magnitude. It does have an associated field energy. So is this radiation ? – Lelouch Sep 08 '16 at 03:24
  • @Lelouch, if I understand your 2nd sentence correctly, you're stipulating that the electric field is given by $\vec E(x,y,z,t) = \vec E,(1 + 10,\Theta(t))$. Is this correct? – Alfred Centauri Sep 08 '16 at 12:26
  • No, I meant that if i suddenly switched its magnitude, then this information will travel at a finite speed c, i.e the field will start altering its magnitude, starting from the source and spreading out, at the speed of c. Is this propagation radiation ? , since this is a 'propagating disturbance' in the current electric field. – Lelouch Sep 08 '16 at 12:32
  • @Lelouch, you first specified that there is a uniform electric field in all space and then state that you suddenly switch its magnitude 10 times. Now, as far as I can tell, it is the uniform electric field in all space and thus, you're stipulating that the magnitude of this uniform electric field in all space suddenly increases by a factor of 10. But then suddenly, in your follow up, you speak of the source. It's not clear to me what you're describing. – Alfred Centauri Sep 08 '16 at 12:49