# B Question about understanding conductors for EM course

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1. May 21, 2016

### Sho Kano

So in conductors, the electrons will distribute themselves to the surface via repulsion forces. But why do we say that the electric field inside is zero? If I put a positive charge inside, clearly it will move in some direction from the electric field of generated from the electrons. Also, are the electrons actually stationary on the surface (i.e. have a net zero force due to each other), or are they moving but have a net effect of zero?

2. May 21, 2016

### houlahound

It appears you are conflating two different experiments/conditions?

3. May 21, 2016

### Staff: Mentor

It is only zero in the electrostatic condition. I.e. given that a system is electrostatic then $J=0$. In a conductor $J=\sigma E$ so $E=0$.

4. May 21, 2016

### Sho Kano

Okay, so the electrons repel each other by repulsion due to each other's E fields, then end up stationary at the surface of the conductor. At this point, the parallel E fields along the surface cancel out to 0, and all that's left are the perpendicular ones. So since there are no electrons below the surface, there is no E field below the surface?

5. May 21, 2016

### Staff: Mentor

The electrons only arrange themselves on the surface if there are excess electrons. IE, if the conductor is negatively charged. If your conductor is an infinitely long cylindrical wire, the e-field from the electrons cancels out in every direction except radially outwards from the surface. Inside the e-field cancels out in all directions, including radially. If it didn't cancel out, you'd have an e-field set up inside the conductor and the charges would move until the e-field is zero.

6. May 21, 2016

### Sho Kano

So in a neutrally charged conductor, the electrons are still attached to the atoms?

7. May 21, 2016

### Sho Kano

They could cancel out like this? i.e. field lines bend upwards- won't the bottom parts contribute to a net downwards field too?

Last edited: May 21, 2016
8. May 21, 2016

### Sho Kano

Alright I understand now; There is a net zero electric field inside a conductor in static equilibrium. If you bring in an electron, it will throw off all the other electrons, and they will rearrange themselves. On a related question, shouldn't the electric field right on the surface of a conductor be zero?

Shouldn't there be a discontinuity at x=r?

9. May 22, 2016

### Staff: Mentor

Most are, yes. Some are free and move about the conductor. The exact number depends on the material the conductor is made out of. While these free electrons are able to move around, their net motion cancels out overall, so you don't see current flow or charged areas in a neutral conductor.

That only works for two charges. There are MANY individual charges. So many that there isn't anywhere for the field lines to "bunch up" like you see in that picture.

I don't think so. On the surface means that it isn't inside the conductor, so the electric field shouldn't be zero there.

10. May 22, 2016

### Sho Kano

So only the excess (net charge) distributes themselves on the surface, while the net charge inside is 0. This is why there is no net E field in the conductor.

The E field inside a neutral conductor is also 0 because there is a net zero charge meaning a net zero field.

11. May 22, 2016

### Staff: Mentor

Again, only in the electrostatic case.

12. May 22, 2016

### Sho Kano

Yes. Looking at this picture though, there seems to be also perpendicular field lines extending into the conductor (from the surface charges)- do these cancel out each other too?

13. May 22, 2016

### Sho Kano

Great!

14. May 22, 2016

### Sho Kano

If I try to apply Gauss's law here, the Gaussian surface will be right on the charges, so there seems to be no enclosed charge?

15. May 22, 2016

### pixel

The "surface" is an idealization - something that is infinitely thin and yet contains any excess charges assumed to be present. What happens exactly at the surface is more of a philosophical question. For instance, if the surface is "right on the charges," then some parts of the charges are within the surface and some parts are outside. That would lead to a transition regime to replace the vertical dashed line in your plot.

16. May 22, 2016

### Sho Kano

How do we calculate the field at a point on the surface? So would Gauss's law still hold, and we would use the radius of the sphere?

17. May 22, 2016

### Staff: Mentor

It is discontinuous on the surface.

18. May 22, 2016

### pixel

Yes, over any reasonable distance scale, but the OP is asking - maybe unnecessarily - about the E field in a small region at the "surface."

19. May 22, 2016

### Sho Kano

Right, it's much clearer now. A quick question, why do we experience electric shocks (i.e. when touching metals) if the human skin is not a great conductor? Lets say I build up a net positive charge from friction, then approach a doorknob. My finger's electric field will induce a negative charge on the doorknob, and when I touch it the electrons flow through my finger, and I feel a shock. Skin is not a great conductor, also the rubber shoes are not great either.

20. May 22, 2016

### Sho Kano

Okay I understand now, but one thing is still incomplete- the field lines would also point inwards as well as outwards. These inward lines add up to a field inside right?

21. May 22, 2016

### Staff: Mentor

Skin and shoes may not be great conductors, but when the voltage between your finger and the doorknob is high enough to ionize the air itself (resistance of around 1013 ohms/mm) the 100,000 ohm resistance of your skin does very little. Also, charge isn't be transferred through your shoes when you shock yourself on a doorknob, you yourself have already accumulated a charge.

Imagine you keep adding charges to your picture above. The field lines coming from each electron would be shoved closer and closer together until they just radiate straight outwards, away from the center of the wire. The field inside the wire is zero, meaning that a charge of either type placed in the middle of the wire feels no net force in any direction. This is just like a slice through a hollow, charged sphere.

22. May 22, 2016

### Sho Kano

What are the situations where charge goes into you and directly to the ground? High enough voltage?
It is still a bit hard to imagine. Yes the field lines would radiate straight outwards, but it seems like the same would happen inwards even if I bunch the charges up (i.e. field lines radiate straight inwards). But since we know the electric to be zero inside a conductor, what happens to these inward lines?

23. May 22, 2016

### Staff: Mentor

No, I could apply 120 volts between your foot and your hand and get current flow. My point was that when you shock yourself your shoes don't afford any protection because your body has already developed a charge.

There aren't any. They would all radiate outwards. Remember that field lines are visualization tools that help us visualize the magnitude and direction of the force exerted on a charged particle if placed in that field. They are not a fundamental explanation in and of themselves. If you place a charged particle inside your conductor and then calculate the net force from every single electron you would find that the net force is zero no matter where you place the charge inside the conductor.

24. May 22, 2016

### Sho Kano

But field lines are accurate representations of the electric field, and the electric field gives an accurate representation of the forces on an charged particle.
I can't visualize why there wouldn't be an electric field inwards. Maybe I have to just downright accept it?

25. May 22, 2016

### Staff: Mentor

Of course. And if the electric field inside the conductor is zero, then there aren't any field lines.

Let me ask you this. In the picture above, what is in between any two field lines? Do charged particles have to be placed on a field line in order to feel a force?