Homework Help: The continuity equation in electromagnetism

1. Apr 1, 2014

physiks

I understand the reasoning behind the equations
SJ.dS=-dQ/dt and thus ∇.J=-∂ρ/∂t.
where the integral is taken over the closed surface S.

The book I'm using sets dQ/dt=0 and ∂ρ/∂t=0 in these cases. I don't understand this very well, as I am possibly blinded by the fact usually I=dQ/dt and so a steady current would mean dQ/dt=constant. I believe that things are different here because we are dealing with the charge flowing out of a volume. I also can't understand the second condition. If somebody could explain them I would be grateful.

I also have seen somewhere else:
For steady currents, ∂J/∂t=0. This seems to say J=constant, so surely I is then constant which disagrees with the above. So how does this agree with the above? Finally I have the condition ∂ρ/∂t for stationary charges, which makes sense. This then leads to ∇.J=0 as above - but is it consistent that ∂ρ/∂t is zero for constant current and stationary charges?

Everything is a bit jumbled up in my head at the moment, so if somebody could explain each of those four conditions physically that would be helpful, thanks.

2. Apr 1, 2014

jaytech

Steady current means the current density is unchanging. ∇$\cdot$$\vec{J}$ = 0 means the current density $\vec{J}$is constant, and therefore for unimpeded flow to occur (steady current), the current distribution in any given volume must be homogeneous (the same everywhere). Now matter where your configuration "starts", $\vec{J}$ will appear the same from every point.

$\frac{dρ}{dt}$ = 0 is analogous to the above, but simply states the number of charges in a given volume is constant, and does not change with time. i.e. Conservation of charge.

3. Apr 1, 2014

physiks

Am I correct in saying that here I does not equal dQ/dt, because Q is no longer a quantity of charge the flows, but the quantity of charge in the volume?

So does a steady current mean J is constant by definition? Or does it mean that such movement of charge must go on forever without charge build up that would stop it? Or both?

Last edited: Apr 1, 2014
4. Apr 1, 2014

jaytech

I = dQ/dt still holds. That equation is correct. It's just easier to think of continuity using $\vec{J}$ and remembering that limA→0 $\frac{I}{A}$ = J.

So really you can say that limA→0 $\frac{dQ/dt}{A}$ = J.

This is similar to $\frac{d}{dt}$( $\frac{dQ}{dA}$ ) which shows that the current density is the amount of charge occupying a cross sectional area changing in time. This is a flow of charge in a volume.

5. Apr 1, 2014

physiks

Hmm but then steady currents -> dQ/dt=0 -> I=0. So current is zero for a steady current. What am I missing?

steady currents -> ∂J/∂t=0 -> J=constant -> I=constant.

Last edited: Apr 1, 2014
6. Apr 2, 2014

physiks

Now the situation is:

I understand the reasoning behind the equations
SJ.dS=-dQ/dt and thus ∇.J=-∂ρ/∂t.
where the integral is taken over the closed surface S.

I mostly understand the conditions for steady currents that can be stated, i.e
J/∂t=0, ∂ρ/∂t=0 and dQ/dt=0.

What I don't understand is how the current I fits into this. I have seen I=-dQ/dt applies to the volume, but I can't see this because:

steady currents -> dQ/dt=0 -> I=0. So current is zero for a steady current.

whilst

steady currents -> ∂J/∂t=0 -> J=constant -> I=constant.

This appears to be a contradiction. Also the current is zero for a steady current which is strange.

Last edited: Apr 2, 2014
7. Apr 2, 2014

jaytech

The current is a constant, as is $\vec{J}$, the current density. Let me explain.

Now the first thing to note is we are dealing with electrostatics and magneto-statics. Electrostatics requires stationary charges, and magneto-statics requires steady currents.

Now, if you take I = $\frac{dQ}{dt}$ at face value, then you are examining a point charge. You are literally saying the rate a single charge moves through time is equal to the current. With a single charge though, you cannot produce a steady current. You require a configuration of charges all moving together with a drift velocity $\vec{v}$. The equation I = 0 = $\frac{dQ}{dt}$ is true for this point charge, and in turn makes an electrostatic problem (since I = 0).

Now it seems you are confusing I = $\frac{dQ}{dt}$ = 0 with magneto-statics, and although it is applicable, the current must be constant and the condition $\frac{dQ}{dt}$ = 0 must be satisfied for each individual charge (constant current = magneto-statics). So how can we do this?

Let's first rewrite I = $\frac{dQ}{dt}$ to this form: I = $\frac{dQ}{dl}$ $\frac{dl}{dt}$ . This now says I = λv, where λ is a line charge density. This is analogous to a wire carrying multiple charges at a drift velocity with magnitude v. Now we can implement the magneto-static condition $\frac{dI}{dt}$ = 0 (or as you said $\frac{dJ}{dt}$ = 0, the only difference is J = Aρv, where A is cross section area, and ρ volume charge density). This would imply that I (or J) is constant. But what of the right side of the equation? The λv? This would be $\frac{d}{dt}$ (λv), which is the same as λa, where a is acceleration. Steady current means there is no acceleration of the charges, it is all constant velocity. So λa = 0.

Well what about the original question? I = $\frac{dQ}{dt}$ = 0? They can't both hold for magneto-statics, can they? Well they can and they do. As I first stated, $\frac{dQ}{dt}$ = 0 represents an electrostatic problem, whereas I = λv represents a magneto-static problem. The difference is that I is no longer defined by a single charge in motion, but by a CONFIGURATION of charges in motion together. This configuration, when moving at constant velocity, is the definition of steady current. That is why for steady current, if we do $\vec{∇}$$\cdot$$\vec{J}$ , or in the case of a wire in the x-direction $\frac{dI}{dx}$ , we must have 0:

1) I = λv
2) $\frac{dI}{dx}$ = $\frac{d}{dx}$ (λv)
3) $\frac{dI}{dx}$ = $\frac{d}{dx}$ ( $\frac{dQ}{dx}$ $\frac{dx}{dt}$ )
4) $\frac{dI}{dx}$ = $\frac{d}{dx}$ ( $\frac{dQ}{dt}$ ) (Constant I for the configuration λ)
5) I = $\frac{dQ}{dt}$ (Dropping the divergence term $\frac{d}{dx}$ gives us a single charge again)

With $\frac{dQ}{dt}$ = 0, I = 0 again.
What happened here? Well by taking the spatial derivative of the current I = λv, we are essentially "breaking" the configuration λ into N point charges, each with current In. As we said before though, a single point charge cannot produce a steady current, so each of these In's must equal zero.

I hope this clarified how current can be constant and $\frac{dQ}{dt}$ = 0 can still hold. Remember, a configuration of charges moving together at drift velocity v implies a constant I (magneto-statics) whilst a single charge moving does not. The I for a single charge must equal zero, such that the problem is electrostatic in nature.

Last edited: Apr 2, 2014
8. Apr 2, 2014

physiks

Thanks for your help, I just can't quite get it to all fit together in my head.

If I have say a cube containing a charge Q and a steady current I flows into one face and out the opposite face, then I can obviously see
dQ/dt=0 as Q is not changing
SJ.dS=0, i.e there is no net flux of the current density into/out of the cube
∂ρ/∂t=0 as the charge density won't change
J/∂t=0 because the current is constant

But I have
SJ.dS=-dQ/dt=I=0 so I=0. But I isn't zero. Is there not a way of visualizing what's going on, as I could probably grasp all the above equations then...

9. Apr 2, 2014

jaytech

You're correct.

Not quite.

S $\vec{J}$$\cdot$d$\vec{S}$ = $\int$S limA→0$\frac{\vec{I}}{A}$$\cdot$d$\vec{A}$

As A trends toward zero in the limit, it cancels the differential dA. This means as you approach the area of a point (point charge), the surface disappears. So again, we are looking at a single point charge, which is why the current is zero (electrostatic).

Remember $\int$S $\vec{J}$$\cdot$d$\vec{S}$ is the same as $\int$V $\vec{∇}$$\cdot$$\vec{J}$ dV. This volume integral clearly shows how you are "breaking" the current density in order to add up the individual point charge contributions.

10. Apr 2, 2014

physiks

So when I am saying 'the current is constant', this is for the magnetostatic current.

When I am saying 'the current is zero', this is from an electrostatic perspective?

11. Apr 2, 2014

jaytech

You just seem to be confusing a single point charge in motion I = $\frac{dQ}{dt}$ with a current density $\vec{I}$ = λ$\vec{v}$. A single point charge in motion is meaningless to examine unless we care about the fields $\vec{B}$ and $\vec{E}$ associated with it. Every example you have described is steady current, meaning the single dimensional current density $\vec{I}$.

They both represent current, but one is the individual contribution of a point charge, and the other is the cumulative contribution of a line of charge.

Realize that by piecing many charges together to create a line charge, the total sum of current is nonzero due to the configuration as a whole moving (drift velocity). Individually, they are not moving. If I add a single charge to the end of the line, it will push the entire configuration. The individual charges within the line do not move relative to the charges around them. Therefore, the configuration moves, whilst the individual components do not. Hence a constant $\vec{I}$ and $\frac{dQ}{dt}$ = 0.

12. Apr 2, 2014

jaytech

Exactly. Remember, E&M is relativistic. The trouble is when you first start E&M, they treat the electric and magnetic fields as separate entities, when they really are not.

Edit: To elaborate, from your viewpoint of the wire, there is a constant current flowing across it. From the viewpoint of the charge however, nothing is moving at all. This way, the charge thinks it is electrostatic, but to the observer (who observes the whole configuration of charges) it is magneto-static.

Last edited: Apr 2, 2014
13. Apr 2, 2014

physiks

So to summarize, for steady currents, we have a constant current I.

If I wanted to write ∫SJ.dS=-dQ/dt=I, I'd be playing with a different definition of I - i.e from the viewpoint of somebody sitting in the volume not aware of what is going on outside. The apparent current inside would be zero.

Anyway, thanks for all of your time, very much appreciated :)

14. Apr 2, 2014

jaytech

Not really a different definition, but yes it is a different viewpoint. You seem to be grasping the relationship, and understanding how both $\frac{dQ}{dt}$ is zero and current constant.

E&M can be very confusing when you first develop these relationships, but the more you work with it and incorporate other parts of physics, the more it makes sense. Awesome job!