# Electric field in a circuit

1. Jul 25, 2014

### tonyjk

Hello,

In an electric circuit, let's say we have a resistor and a generator(Voltage source). The electric field inside the resistor is due to a charge distribution inside the resistor or from the propagation of the electric field coming from the voltage source?

Thank you

2. Jul 25, 2014

### atyy

It is due to the electric field generated by the voltage source. The resistor itself is uncharged (in normal conditions).

3. Jul 25, 2014

### tonyjk

Thank you. Just a last question : in an inductor, the conservative electric field that compensate the non-conservative one is coming from what?

4. Jul 25, 2014

### atyy

I don't understand the question. Can you point me to a reference so I can see the context?

5. Jul 25, 2014

### tonyjk

Here's a reference http://www.physics.ucf.edu/~roldan/classes/Chap30_PHY2049.pdf (page 5-6). There's the induced electric field(non-conservative) and a conservative electric field inside an inductor

6. Jul 26, 2014

### tonyjk

please is there anyone who can help

7. Jul 26, 2014

### vanhees71

I don't know, at which level these lectures are, but the manuscript seems not to be very clear in defining things.

The correct way to treat electromagnetic problems are Maxwell's equations in differential form. Here, it's mostly Faraday's Law that is important:
$$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0.$$
As you see, if there is a time-dependent magnetic field, the electric field is not conservative, because its curl doesn't vanish.

The integral law, which seems to be applied in the lecture notes (transparencies), is derived by the use of Stokes's integral theorem. To that end integrate Faraday's Law over an arbitrary surface $F$ with boundary $\partial F$. The orientation of the surface elements $\mathrm{d}^2 \vec{F}$ and that of the line elements $\mathrm{d} \vec{r}$ of the boundary curve are according to the right-hand rule. Then you have
$$\int_F \mathrm{d}^2 \vec{F} \cdot (\vec{\nabla} \times \vec{E}) = \int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{1}{c} \int_{F} \mathrm{d}^2 \vec{F} \cdot \partial_t \vec{B}.$$
Now comes a very subtle point! We want to get the time derivative out of the integral. This can be done in the naive way only, if your surface and its boundary are stationary, i.e., if they are not time dependent. Then you have
$$\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_F \mathrm{d}^2 \vec{F} \cdot \vec{B}.$$
In this (and only in this!) case the electromotive force is given by the left-hand side, and the right-hand side is the time derivative of the magnetic flux, i.e., you have
$$\mathcal{E}=\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{E}, \quad \Phi=\int_{F} \mathrm{d}^2 \vec{F} \cdot \vec{B},$$
and the integral form of Faraday's Law can be written as
$$\mathcal{E}=-\frac{1}{c} \frac{\mathrm{d} \Phi}{\mathrm{d} t}.$$
If your surface is moving, there's an additional term in the definition of the electromotive force, namely
$$\mathcal{E}=\int_{\partial F} \mathrm{d} \vec{r} \cdot \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right ),$$
where $\vec{v}=\vec{v}(t,\vec{r})$ is the velocity of the point $\vec{r}$ on the boundary curve.

For a derivation, see the Wikipedia

I hope, now the principles are a bit more clear. If not, ask again!

8. Jul 26, 2014

### DrZoidberg

Take a look at those documents.
http://www.phy-astr.gsu.edu/cymbalyuk/Lecture16.pdf [Broken]
http://www.if.ufrj.br/~dore/Fis3/Coombes-AJP.pdf
http://www.ifi.unicamp.br/~assis/Found-Phys-V29-p729-753(1999).pdf

So in short, there is electric charge on the surface of a resistor (but not inside) as long as there is a current.
That charge is the main source of the electric field inside. It's not the only source since there are also charges on other parts of the circuit which also produce fields.
http://www.physics.ucf.edu/~roldan/classes/Chap30_PHY2049.pdf
is talking about on page 5 where it says: "E produced by charges distributed around circuit is conservative Ec".

Last edited by a moderator: May 6, 2017
9. Jul 26, 2014

### tonyjk

So the electric field inside a resistor is due to the electric charge on the surface and not from the propagation of electric field from the voltage source? We have now 2 different ideas i am a little bit confused

Last edited by a moderator: May 6, 2017
10. Jul 26, 2014

### tonyjk

We can say that the voltage source maintain the charge distribution on the surface of a wire or resistor right? and from where these charges are coming?

Last edited: Jul 26, 2014
11. Jul 26, 2014

### atyy

When the electric field does not change with time, and charges are stationary, the electric field is conservative and can be fully described by the electric potential.

However, in a circuit, for current to "flow through" a capacitor or inductor, then current must be time varying, and so the electric field is also time-varying, and it is not conservative. This means that strictly speaking, we cannot use the electric potential to describe a circuit exactly.

However, the electric potential can still be used to approximately describe time varying currents such as alternating currents in a circuit, provided the changes in current are relatively slow. This is why we still use the electric potential in circuit theory, even though strictly speaking there is energy being radiated away from the capacitors and inductors.

Last edited: Jul 26, 2014
12. Jul 26, 2014

### DrZoidberg

The term "non conservative electric field" usually doesn't refer to the loss of energy due to em radiation. It means circular fields induced by changing magnetic fields.
There is a great MIT lecture about it here

It is all explained in the pdfs I linked.

Last edited by a moderator: Sep 25, 2014
13. Jul 26, 2014

### atyy

Yes, that's right, so let me clarify. The loss of energy by radiation I referred to indicates that the electromagnetic field is changing with time. This time variation means that the electric field is non-conservative, since only an electric field that has no time variation is conservative and exactly describable by a potential. By using a potential in circuit theory, we ignore the non-conservative aspect of the field. For example, in circuit theory, one can have oscillations that don't die away in a closed circuit consisting of an inductor and capacitor in series. In the full electromagnetic theory, this is not possible, because the inductor and capacitor will radiate.

Formally:

1) Only an electric field that does not change with time is conservative.
2) If an electric field is conservative, it can be exactly described by a potential.
3) The conservative nature of the potential is expressed in Kirchoff's second law.
4) Kirchoff's second law: the sum of potential differences taken around a closed circuit is zero.

5) When a current is changing through a capacitor or inductor, the situation is time varying and no longer conservative.
6) In particular, there is radiation that carries energy away from the capacitor or inductor.
7) However, if the time-variation is slow, the energy loss by radiation is very small and approximately zero, so we can still use a potential.
8) In particular, we can still use Kirchoff's second law as a very good approximation, as if the situation is conservative.

9) The important condition that defines a conservative field in circuit theory is that Kirchoff's second law holds. In the full electromagnetic theory, this is expressed as "the line integral of the electric field around a closed loop is zero" or the "curl of the electric field is zero". But that's just for reference when you get to Maxwell's equations.

Last edited by a moderator: Sep 25, 2014
14. Jul 26, 2014

### tonyjk

Okay Thank you I believe these charges come from the surrounding air

15. Jul 26, 2014

### DrZoidberg

16. Jul 26, 2014

### tonyjk

http://www.ifi.unicamp.br/~assis/Found-Phys-V29-p729-753(1999).pdf(page 10).
What I am not understanding is that the surface charges of the resistor are driving the current wich is mainly composed of electrons' resistor(negative charges). so we have many types of charges in a resistor?

17. Jul 26, 2014

### tonyjk

I have another "related" question : when we turn off a switch in a circuit, there's an electric arc, ionisation of the surrounding air due to a high electric field. my question is this high electric field is due to surface charges or from the non-conservative electric field coming from an inductor in the circuit?

18. Jul 26, 2014

### atyy

That's an interesting question. I don't know the answer and hope someone else will contribute. My guess is that the arc during switching is mainly caused by the very fast change in the current. If there is inductance in the wires, this will create a non-conservative electric field, which if high enough, could result in an arc. (That's a guess)

Here are some links that indicate that an arc can be due to the large electric field caused by a switching transient and inductance in the circuit:
http://see.msfc.nasa.gov/publications/CR-2002-211839.pdf [Broken] (section 8.0)
https://vet.bme.hu/sites/default/files/tamop/vivem174en/out/html/vivem174en.html (Fig. 2.22-2.23)

Last edited by a moderator: May 6, 2017
19. Jul 27, 2014

### tonyjk

does anyone have a link or something about it? i am not finding anything on the internet

20. Jul 27, 2014