MHB How do you find the direction of a displacement current?

[Ackbach]

We know that the so-called displacement current is defined as
$$i_d=\varepsilon_0 \, \frac{\partial\Phi_e}{\partial t}.$$
Like regular current which is the movement of charges, $i_d$ has a direction, even though it's technically a scalar. How do we find its direction?

 

[I like Serena]

We know that the so-called displacement current is defined as
$$i_d=\varepsilon_0 \, \frac{\partial\Phi_e}{\partial t}.$$
Like regular current which is the movement of charges, $i_d$ has a direction, even though it's technically a scalar. How do we find its direction?

Hi Ackbach,

From wiki, the displacement current density is:
$$\boldsymbol{j}_d = \pd {\boldsymbol{D}} t = \varepsilon_0 \frac{\partial \boldsymbol{E}}{\partial t} + \frac{\partial \boldsymbol{P}}{\partial t}$$

If we pick some surface, through which we want to know the displacement current, this is:
$$i_d=\iint \boldsymbol{j}_d \cdot d\boldsymbol S
= \iint \Big(\varepsilon_0 \frac{\partial \boldsymbol{E}}{\partial t} + \frac{\partial \boldsymbol{P}}{\partial t}\Big) \cdot d\boldsymbol S
= \varepsilon_0 \iint\frac{\partial \boldsymbol{E}}{\partial t}\cdot d\boldsymbol S + \iint \frac{\partial \boldsymbol{P}}{\partial t} \cdot d\boldsymbol S
$$
For a fixed surface in a medium with constant polarization (such as vacuum), it simplifies to:
$$i_d = \varepsilon_0 \frac{\partial \Phi_e}{\partial t}$$

By doing this, we have "lost" the direction.
If we want to get the direction back, we can measure $i_d$ through 3 perpendicular surfaces of unit size.
The corresponding results are the components of the vector with respect to the normals of those surfaces.
Alternatively, we can search for the surface that has the greatest $i_d$. Its normal is the direction.

Btw, I think this topic belongs in Other Advanced Topics, so I've moved it there.