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A common exercise in non-stationary electromagnetism is to find the electric and the magnetic field generated by a capacitor with round plates, if the potential difference between the plates varies in time (typically $V=V_0 sin( \omega t)$ or $V= \alpha t$).

While finding the fields internal to the capacitor is pretty easy, I was struggling with those on the outside. $B$ is given instant per instant by Biot-Savart law and varies in time, so exists an electric field following the third Maxwell equation; However, I couldn't find an expression for $E$ without evident absurdities.

I thought that maybe with some approximations (quasi-stationarity?) such fields could be ignored. Some hints?

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  • $\begingroup$ How are you calculating the electric and magnetic field? directly? $\endgroup$
    – Hydro Guy
    Sep 12, 2014 at 15:34
  • $\begingroup$ $B$ directly, of the electric field I only know $\nabla \times \vec{E} = - \frac{\partial \vec{B}} {\partial t}$. $\endgroup$
    – Fabio
    Sep 12, 2014 at 15:43
  • $\begingroup$ how are you calculating the Magnetic field outside? $\endgroup$
    – Hydro Guy
    Sep 12, 2014 at 15:45
  • $\begingroup$ What expression did you get, what were the absurdities you talk about? $\endgroup$ Sep 12, 2014 at 15:50
  • $\begingroup$ from Biot-Savart law, in cylindrical coordinates $\vec{B}$ is directed toward $\theta$, and $B(r)= \frac{\mu _0 I} {2 \pi r}$. The same in the area between the plates outside the capacitor, using displacement current instead of $I$ $\endgroup$
    – Fabio
    Sep 12, 2014 at 15:53

2 Answers 2

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The answer depends upon how complicated you want your answer to become. An example of a very similar question is posed in John D. Jackson's 3rd Edition of his Classical Electrodynamics, problem 6.14. There are a couple of ways to approach the problem.

First approach:
Assume that the current to the capacitor is given by: $$ I\left( t \right) = I_{o} e^{-i \omega t} $$ then the charge density, $\sigma$, on one plate at any given time is given by: $$ \sigma \left( t \right) = \frac{ i I\left( t \right) }{ \omega \pi a^{2} } $$ assuming the plates have a radius a and separation d. If we ignore the second plate for the moment (and dielectrics), we can show (using Gauss' law) that the electric field is given by: $$ \oint dA \ \mathbf{E} \cdot \hat{\mathbf{n}} = \frac{ 1 }{ \varepsilon_{o} } \int dA \ \sigma $$ where, in the simplest approximation (i.e., uniform $\sigma$, $\varepsilon$ = $\varepsilon_{o}$, and d $\ll$ a), then E = $\sigma$/2$\varepsilon_{o}$ $\hat{\mathbf{n}}$. A similar discussion was already posted here.

A more general result, but still simple is to consider the same idea and find the electric field on the axis of symmetry of the plate at some distance z away from the plate. Then the the above equation could be rewritten in differential form as: $$ dE = \frac{ z dq }{ 4 \pi \varepsilon_{o} \lvert \mathbf{r} - \mathbf{r}' \rvert^{3} } \\ = \frac{z \ \sigma \ dA}{ 4 \pi \varepsilon_{o} \left( r^{2} + z^{2} \right)^{3/2} } \\ = \frac{ 2 \pi \ z \ r \ \sigma \ dr }{ 4 \pi \varepsilon_{o} \left( r^{2} + z^{2} \right)^{3/2} } $$ Then one could add another plate and find the superposition of the two fields.

More general approach:
Assume a current is introduced at the center of the plate (i.e., r = 0), then we can write a continuity equation for the charge density as: $$ \partial_{t} \ \sigma + \nabla \cdot \boldsymbol{\kappa} = \frac{ I\left(t\right) }{ 2 \pi r } \delta \left( r \right) $$ where $\boldsymbol{\kappa}$ is a surface current density, $\delta \left( r \right)$ is the Dirac delta function, and $\sigma$ is a time-dependent surface charge density. In the idealized case, Ampere's law will go to: $$ \hat{\mathbf{n}} \times \mathbf{B} = \mu_{o} \boldsymbol{\kappa} $$

The trick is then determining $\sigma$ and $\boldsymbol{\kappa}$ as functions of r, t, a, and $\omega$. The end result is that both depend upon Bessel functions, as you might imagine from the cylindrical geometry.

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  • $\begingroup$ This is correct but, if I got you right, these are expressions for the electric field inside the capacitor; I was looking for the electric field outside. $\endgroup$
    – Fabio
    Sep 12, 2014 at 16:17
  • $\begingroup$ Ah yes, sorry about that. You could look at one plate plus the wire feeding the plate as a system. The wire contributes the typical magnetic field proportional to (I/r) and the plate contributes ~K, both being time-dependent. The electric field would then have two parts, one from the time-varying charge density on the plate (Gauss' law) and one from the time-varying magnetic fields (Faraday's law). If you assume the plates are close enough and large enough, then the latter term will dominate and the former can be neglected. $\endgroup$ Sep 12, 2014 at 16:32
  • $\begingroup$ Thanks! How can I show that such term may be neglected? $\endgroup$
    – Fabio
    Sep 12, 2014 at 16:36
  • $\begingroup$ One can't neglect these terms. Outside of the capacitor you would have either a finite size loop producing something of a dipole field or an infinite antenna. Unless you are restricting yourself to a special volume, where the problem can be simplified, you would have to solve for the complete radiation field (which, for the infinite antenna, would lead to a divergent radiative power, of course). $\endgroup$
    – CuriousOne
    Sep 13, 2014 at 0:16
  • $\begingroup$ That's the answer I was looking for! I'm not practical with radiation, can you give me a hint on how to formalize the problem? (If it's not excessively complex, otherwise I can stop here) $\endgroup$
    – Fabio
    Sep 13, 2014 at 10:17
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I think you are making a mistake in assuming the sum of electric field has to follow Maxwell's laws here.

In this case you have simple electric field in the wires (radial, along wires) and "induced" electric field between capacitor plates.

It is well established that sum of induced electric field over closed loop need not be zero, hence when you sum over all the fields over a loop you seem to be getting non-zero answer which is acceptable here.

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  • $\begingroup$ Yes, but you are referring to the electric field inside the capacitor (by the way, the electric field inside, resulting from the charge deposited on the plates, is conservative). My question regarded the field outside. $\endgroup$
    – Fabio
    Sep 12, 2014 at 16:55
  • $\begingroup$ To complete the loop you need to cross the capacitor and if you don't consider the field between the capacitor you can't complete the loop and then can't apply the law. $\endgroup$ Sep 12, 2014 at 16:58
  • $\begingroup$ Ok, so how can I find the field outside? $\endgroup$
    – Fabio
    Sep 12, 2014 at 17:03
  • $\begingroup$ you can use $i=\frac{n e^2 A \tau E}{m}$, keep in mind outside the capacitor, the changing electric field is causing the magnetic field and not the other way around. If you try to find out electric field by differentiating the function of B you would end up with a different induced electric field, which may or may not be what you are looking for. $\endgroup$ Sep 12, 2014 at 17:18
  • $\begingroup$ I was indeed trying to find the electric field induced by the magnetic field $\endgroup$
    – Fabio
    Sep 12, 2014 at 21:19

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