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?
 A: Let me give a very simple example for extensive and intensive quantities crudely. For instance if you stick two objects $30 \;\text{kg}$ each, you get an object with $60 \;\text{kg}$. This is an extensive property. But if both objects have the temperature of $30 \;\text{K}$, when you stick them then temperature is $30 \;\text{K}$ and not $60 \;\text{K}$. This is an intensive property. Now, electric field is a vector. When you combine two electric fields at a point in space, rather than summing magnitudes directly, you need to take direction into consideration (30+30 doesn't mean 60 for vectors). Therefore intuitively electric field is intensive. Electrical potential energy would be extensive.
A: 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. 
