# How to prove that $\vec{E}$ is an intensive property?

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?

• I can make a good guess as to what you mean, but you really should define your variables – Aaron Stevens Jul 27 '19 at 23:52

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 $$\sigma$$.
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.