The Electric Displacement, ##\mathbf D##

[Philip Wood]

TL;DR

Why isn't it defined so that it has the same units as $\mathbf E$?

We define $\mathbf D$ as $\mathbf D=\epsilon_0\mathbf E + \mathbf P$, in which $\mathbf E$ is the electric field strength and $\mathbf P$ is the polarisation.

Would it not be more convenient for $\mathbf D$ to be defined in such a way that it had the same units as $\mathbf E$. In other words as $\mathbf D=\mathbf E + \frac 1 {\epsilon_0} \mathbf P$ ?

 

[Dale]

That is just a matter of the choice of units. In Heaviside Lorentz units all three quantities have the same units which is even more convenient than moving the constant around.

https://en.wikipedia.org/wiki/Heaviside–Lorentz_units

 

[Meir Achuz]

TL;DR: Why isn't it defined so that it has the same units as $\mathbf E$?

We define $\mathbf D$ as $\mathbf D=\epsilon_0\mathbf E + \mathbf P$, in which $\mathbf E$ is the electric field strength and $\mathbf P$ is the polarisation.

Would it not be more convenient for $\mathbf D$ to be defined in such a way that it had the same units as $\mathbf E$. In other words as $\mathbf D=\mathbf E + \frac 1 {\epsilon_0} \mathbf P$ ?

In Gaussian units, which I use, $\mathbf E$=q$\mathbf r$/r^3
& $\mathbf D=\mathbf E+4\pi\mathbf P$.