Showing Charge & Currents in Vol: Proving $\mathbf{p}$ Dipole Moment

[dRic2]

Homework Statement

For a configuration of charges and currents confined within a volume $V$, show that
$$\int_V \mathbf J d \tau = \frac {d \mathbf p}{dt}$$
where $\mathbf p$ is the total dipole moment.

Homework Equations

...

The Attempt at a Solution

I have one question: since the configuration of charges and currents is confined within a volume can I assume
$$∇ \cdot \mathbf J =0 $$
?
I mean, if there is no "in" nor "out", $\frac {\partial \rho}{\partial t}$ should be zero, right ?

 

[Delta2]

Not sure , I think you can only assume that if S is the boundary surface of V, then $\oint_S \mathbf{J\cdot dS}=0$ which by divergence theorem means that $$\int_V \nabla \cdot \mathbf{J}dV=0$$ but this last equality holds only for volume V and not for every subvolume of V so this does not imply that $\nabla \cdot \mathbf{J}=0$.

The whole point is that the charges and currents are not confined to a random subvolume of V so there can be flux in or out through a subvolume of V.

 

[dRic2]

Ok, I've given up on this one...

The exercise came with a hint: "Evaluate $\int_V ∇ \cdot( x \mathbf J) d \tau$ "

Here as I tried (but it is wrong):
$$ \int_V ∇ \cdot ( x \mathbf J) d \tau = \int_V x ∇ \cdot \mathbf J d \tau + \int_V \mathbf J \cdot ∇ x d \tau$$
then I said "well $\int_V x ∇ \cdot \mathbf J d \tau$ should be zero because..." but it is wrong.

I checked the solutions and the author says:
$$\int_V ∇ \cdot ( x \mathbf J) d \tau = \int_A x \mathbf J \cdot d \mathbf a$$
is zero because since $\mathbf J$ is entirely inside $V$, it is zero on the surface.

What ? I do not understand... How can he claim that from $\int_A \mathbf J \cdot d \mathbf a = 0$ follows $\int_A x \mathbf J \cdot d \mathbf a = 0$ ?

PS: how do you write the symbol for the integral over a closed surface ?

 

[Delta2]

Ok, I've given up on this one...

...

What ? I do not understand... How can he claim that from $\int_A \mathbf J \cdot d \mathbf a = 0$ follows $\int_A x \mathbf J \cdot d \mathbf a = 0$ ?

PS: how do you write the symbol for the integral over a closed surface ?

It is not only that $\int_A \mathbf J \cdot d \mathbf a = 0$ at the boundary surface A, it is also $J=0$ at the boundary surface. Because the current and charges are confined within the volume V, the current density J is zero everywhere at the boundary surface A of V. If there was current density $\mathbf{J}\neq 0$ somewhere at the boundary surface A, this means there is flow of charge into or outside the volume V.

The symbol for the integral over a closed surface is \oint. Generally you can check the tex commands used by right clicking on an equation and choose view as tex commands and this will open an window with the Latex commands used to make that equation.

 

[dRic2]

If there was current density J≠0J≠0\mathbf{J}\neq 0 somewhere at the boundary surface A, this means there is flow of charge into or outside the volume V.

Aahhhhhhhhhhhhhhhhh Now I get it! I used to think "well, there could be some charge moving around the surface"... but that wouldn't be a "volumetric" current density: that would be a surface current density! Now everything fits! Thank you :)

 

[Delta2]

Ok fine but I have to correct my self a bit, it is not exactly that $J=0$ at the boundary surface but that the component of $J$ normal to the surface element dA is zero, everywhere in the boundary surface (otherwise there would be flow of charge into or outside the volume V), so the dot product $\mathbf{J}\cdot\mathbf{da}$ is zero everywhere in the boundary surface.

 

[dRic2]

If I recall correctly $\mathbf J = \frac {d \mathbf I}{d a_{⊥}}$ so it has to be normal to the boundary (no need to specify further).

 

[kuruman]

If I recall correctly $\mathbf J = \frac {d \mathbf I}{d a_{⊥}}$ so it has to be normal to the boundary (no need to specify further).

I have never seen this definition. Current $I$ is a scalar, not a vector. The total current through a surface $S$ is $$I=\int_S \vec J \cdot \hat n~dS$$ where $\hat n~$ is the local normal to the surface. This is another way of saying what @Delta2 posted in #6.

 

[dRic2]

I adopted the definition given in Griffiths' "Introduction to electrodynamics": (pag 208) the current $\mathbf I$ is defined as a vector, and from that follows the definition of $\mathbf J$ that I posted (pag 210).

 

[kuruman]

I adopted the definition given in Griffiths' "Introduction to electrodynamics": (pag 208) the current $\mathbf I$ is defined as a vector, and from that follows the definition of $\mathbf J$ that I posted (pag 210).

I see. I prefer to stick with $\vec J$ because that's what appears in Maxwell's equations, Ohm's law in vector form and the continuity equation.

 

[dRic2]

I see. I prefer to stick with →JJ→\vec J because that's what appears in Maxwell's equations, Ohm's law in vector form and the continuity equation.

I don't see any contradictions. The definition of $\mathbf J$ is the same, the only things that changes is the definition of $I$, but $I$ does not appear in Maxwell equations, nor continuity equation. I don't know about Ohm's law for now.

 

[PeroK]

If I recall correctly $\mathbf J = \frac {d \mathbf I}{d a_{⊥}}$ so it has to be normal to the boundary (no need to specify further).

This question came up recently. If the current is contained in some volume, then you can argue that the current on the boundary is either zero or tangential to the boundary, hence $\vec{J}.\vec{da} = 0$.

But, you can always take the volume over which you are integrating to be larger than the volume that contains the current. That adds no extra charge or dipole moment and ensures that the current density is zero on the boundary.