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In my homework I had a question to sort a few variables into intensive properties and extensive properties. I wrote that $\vec{E}$ (electric field) is an extensive property, thinking of a situation with a uniform electric field. My answer was marked as wrong.

How does one prove that $\vec{E}$ is intensive?

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  • $\begingroup$ I can make a good guess as to what you mean, but you really should define your variables $\endgroup$ – Aaron Stevens Jul 27 '19 at 23:52
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Even considering a situation with a uniform field, it is not possible to deduce that the electric field $ \bf E$ is extensive. Indeed, the uniform field inside a parallel plane condenser depends only on the surface charge density $\sigma$.

If we put together two such finite condensers (with a surface area and distance between the faces such that rim effects are negligible) either joining them in series (removing the ovelapping opposite charged surfaces with zero work) or in parallel, the internal field remains the same, while the internal energy doubles because we have doubled the volume where the uniform field is.

In a more formal way, we could start from the adiabatic work performed on the system when some charge is varied. It is possible to show (see for instance Landau&Lifshitz textbook on Electrodynamics of continuous media) that the differential of the internal energy $U$ can be written as $$ dU = \int {\bf E}{\delta \bf D}dV $$ where $\delta \bf D$ i the variation of the dielectric induction field ${\bf D}$ induced by a variation of the system charge $\delta q$.

In the case of uniform fields $\bf E$ and $\bf D$, the above expression can be written: $$ dU = V {\bf E}{\delta \bf D}. $$ From such an expression we can see that the if the internal energy has to be a homogeneous function of degree one of all its extensive variables, the presence of the $V$ factor requires that both the $\bf E$ and $\bf D$ field would be intensive quantities. In agreement with the electrostatic analysis sketched above.

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