I Electric continuity equation

1. Feb 28, 2017

jonjacson

Divergence of J= - ∂ρ/∂t (equation 1)

Where ρ is the density of electric charges/ volume

J= the current density = Amperes/m2

I understand that if the divergence is not zero, the rate of change of the amount of charge is changing inside a closed control surface. But I am reading the book from Purcell, and he says:

"In a region where the current is steady div J = 0 (I understand this, simply for a steady current the same amount of charges enters and exits from the surface) together with equation 1, this tells us that the charge density is zero within that region".

I don't understand the last part.

If divJ=0, it means the same amount of charge enters and exists the surface, but there is charge inside the surface right? It does not mean the charge inside the surface is zero. Is this correct?

What I think is:

Divergence of E equal to zero ----> the net charge inside the closed surface is zero
But divergence of current J equal to zero---> the current density is constant

Divergence of J= 0 means the charge density is zero within that region.

What I don't understand is, if the same amount of charge is entering and exiting from the surface the div J will be zero but still some charge density different from zero will be inside the surface.

What do you think?

2. Feb 28, 2017

Khashishi

What's equation 1?

3. Feb 28, 2017

jonjacson

Continuity equation:

Div J = - ∂ρ/∂t

4. Feb 28, 2017

Khashishi

I agree with you. You can have charge with divergence of current = 0. Maybe this is only for a special case. Is there any context that you left out?

5. Feb 28, 2017

jonjacson

Well it is chapter 4 of Purcell's classic book. He is explaining conductivity and Ohms law. I don't know what I am missing.

6. Feb 28, 2017

Staff: Mentor

No, it is the continuum version of Ohms law. Go back and check the text. I think you accidentally misread what he is saying

7. Mar 1, 2017

vanhees71

This is not Ohm's Law but the continuity equation, which states that electric charge is conserved. To see this, integrate the equation
$$\partial_t \rho=-\vec{\nabla} \cdot \vec{j}$$
over some volume which is at rest. Then you get, using Gauss's integral theorem
$$\frac{\mathrm{d}}{\mathrm{d} t} \int_V \mathrm{d}^3 \vec{x} \rho=\dot{Q}_V=-\int_{\partial V} \mathrm{d}^2 \vec{A} \cdot \vec{j}.$$
The left-hand side is the time derivative of the charge contained in the volume $V$. The equation says that the change of the charge is due to the flow of charges through the surface $\partial V$ of the volume. The sign comes from the convention that the surface-normal vectors are pointing out of the surface. Thus a positive $\mathrm{d} \vec{A} \cdot \vec{j}$ means that charge is flowing out of the volume, i.e., within the time $\mathrm{d} t$ the total charges inside the volume decreases.

Ohm's Law in local form is
$$\vec{j}_{\text{cond}}=\sigma \left (\vec{E} + \frac{\vec{v}}{c} \times \vec{B} \right).$$
For usual situations in everyday electrical engineering you can neglect the magnetic term, because the drift velocity $\vec{v}$ of the charges in wires is very small (at the order of millimetres per second), i.e., you have
$$\vec{j}_{\text{cond}}=\sigma \vec{E},$$
where $\sigma$ is the electric conductivity of the medium.

8. Mar 1, 2017

Staff: Mentor

Right, but that isn't what is confusing the OP. Purcell uses the continuity equation and Ohms law to show that the charge inside a conductor is 0 under steady conditions. The OP is getting confused because he misread the proof and is trying to prove it only from the continuity equation.

9. Mar 4, 2017

vanhees71

Purcell again, I see... SCNR.