# Energy Flow in Coaxial Cable with Linear Free Charge Density

## Homework Statement

An infinitely long cylindrical capacitor with inner radius a and outer radius b carries a free charge per unit length of ##\lambda_{free}##. The region between the plates is filled with a nonmagnetic dielectric of conductivity ##\sigma##. Show that at every point inside the dielectric the conduction current is exactly compensated by the displacement current so that no magnetic field is produced in the interior. Find the rate of energy dissipation per unit volume at a point a distance ##\rho## from the axis. Show that the total rate of energy dissipation for a length l is equal to the rate of decrease of electrostatic energy of the capacitor.

## Homework Equations

Maxwell Equations, Ohm's Law, Energy of Capacitor = ##\frac{1}{2} CV^2##

## The Attempt at a Solution

Before getting far into the problem I am confused on two points. I can understand where the conduction current comes from, that's just applying Ohm's law inside the dielectric, but I cannot see where the displacement current comes from or how to evaluate the energy dissipation if the magnetic field is zero inside the dielectric.

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kuruman
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Gold Member
I interpret this problem as a "leaky" capacitor. You charge the central conductor to ##\lambda_{free}##, remove the charging battery and allow the positive charge to leak out to the other conductor and neutralize the negative charge. This gives you a time-varying ##\vec{D}## and hence a displacement current between the conductors. Energy is dissipated by Ohmic losses in the dielectric until the capacitor is completely discharged.

Thank you for the reply. I've attempted the question up to the energy dissipation part now. The problem stated that the central conductor has ##-\lambda_{free}##, so applying Gauss' Law to a cylinder contained in the dielectric at a radius r and with a length L the electric field inside the dielectric is given by:
## E = \frac{-\lambda_{free}}{2\pi r}##.
This induces a conduction current density according to Ohm's Law of ##J = \sigma E = \frac{-\sigma \lambda_{free}}{2\pi r}##.
We can turn this into a differential equation for ##\lambda_{free}## by noting that ## J (2\pi r L)= - \frac{dQ}{dt}## where ##Q = -\lambda_{free} L##.
Therefore ## \frac{d}{dt} (-\lambda_{free} L) = \sigma \lambda_{free} L##, implying that ##\lambda_{free} (t) = \lambda_{free} e^{-\sigma t}##.
So ##\frac{\partial E}{\partial t} = \frac{\sigma \lambda_{free}}{2\pi r}##, and the conduction and displacement currents indeed cancel giving zero magnetic field inside the dielectric.

Am I on the right track with the above work? Additionally I am still at a loss for how to find energy dissipation

kuruman
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You are on the right track, but you need to be careful because of the dielectric. Use ##\vec{D}## in Gauss's Law. Use ##\vec{D}=\epsilon \vec{E}## to find E in Ohm's Law.

Right, electric displacement is used for dielectrics. Correcting that error gives me:
## E = \frac{-\lambda_{free}}{2\pi \epsilon r}##, so the DE for the charge density becomes ##\frac{d}{dt} \lambda_{free} = -\frac{\sigma}{\epsilon} \lambda_{free}##, and the displacement current with the polarization current contribution still cancels the conduction current and it can be concluded that the magnetic field vanishes inside the dielectric from Ampere's Law.

kuruman
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I think you got it now.

So I've worked out the rest of the question and I can get the sensible answer that the rate of energy dissipation per unit volume, ## J \cdot E##, equals the rate of decrease of the electrostatic energy of the capacitor, ## - \frac{d}{dt} [\frac{1}{2} E \cdot D]##, if the Joule heating term only contains the conduction current. But I don't see physically why this should be so.

kuruman