B Question about understanding conductors for EM course

[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?

 

[houlahound]

It appears you are conflating two different experiments/conditions?

 

[Dale]

But why do we say that the electric field inside is zero?

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 $E=0$.

 

[Sho Kano]

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 $E=0$.

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?

 

[Drakkith]

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.

 

[Sho Kano]

The electrons only arrange themselves on the surface if there are excess electrons. IE, if the conductor is negatively charged.

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

 

[Sho Kano]

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

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

 

[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?

 

[Drakkith]

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

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.

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

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.

On a related question, shouldn't the electric field right on the surface of a conductor be zero?

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.

 

[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.

 

[Dale]

Again, only in the electrostatic case.

 

[Sho Kano]

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.

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?

 

[Sho Kano]

Again, only in the electrostatic case.

Great!

 

[Sho Kano]

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.

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?

 

[pixel]

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?

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.

 

[Sho Kano]

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.

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?

 

[Dale]

It is discontinuous on the surface.

 

[pixel]

It is discontinuous on the surface.

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

 

[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? Let's 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.

 

[Sho Kano]

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.

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?

 

[Drakkith]

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? Let's 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.

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.

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?

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.

 

[Sho Kano]

Also, charge isn't be transferred through your shoes when you shock yourself on a doorknob, you yourself have already accumulated a charge.

What are the situations where charge goes into you and directly to the ground? High enough voltage?

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.

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?

 

[Drakkith]

What are the situations where charge goes into you and directly to the ground? High enough voltage?

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.

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 (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?

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.

 

[Sho Kano]

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.

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.

There aren't any. They would all radiate outwards.

I can't visualize why there wouldn't be an electric field inwards. Maybe I have to just downright accept it?

 

[Drakkith]

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.

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

I can't visualize why there wouldn't be an electric field inwards. Maybe I have to just downright accept it?

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?

 

[Sho Kano]

Do charged particles have to be placed on a field line in order to feel a force?

The picture is just a simplification, there are field lines everywhere in space. But if we bunch up a whole lot of them in a line, there will be a net upwards field- And also a net downwards field. The situation doesn't seem to change if we bunch them up along a circle. So in the situation of a circle, we are left with field lines pointing towards the center of the circle.

 

[pixel]

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? Let's 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.

Charge builds up on your skin. Before touching the metal, it jumps across the air making a spark. It's less prevalent in humid weather as it's not as easy for the air to break down.

 

[davenn]

It's less prevalent in humid weather as it's not as easy for the air to break down.

That is not the reason
The reason is that on humid days things don't hold a charge so easily as the humid air allows it to leak away
rather that build upDave

 

[Drakkith]

The picture is just a simplification, there are field lines everywhere in space. But if we bunch up a whole lot of them in a line, there will be a net upwards field- And also a net downwards field. The situation doesn't seem to change if we bunch them up along a circle. So in the situation of a circle, we are left with field lines pointing towards the center of the circle.

Well, you can see for yourself from your picture that the lines do indeed bunch up and curve away from the lines of other like charges. We can add as many lines to each charge as we like and this will still be true. If we start adding more charges into our picture, putting them in the shape of a circle, you'll find that the field lines are bent even more strongly. Add enough charges, so many that we can stop thinking about them as discrete objects and more like a continuous circle of charge, and you'll find that the field lines no longer go inward at all. And how could they? If we have so many charges that we can treat them as continuous, any field line inside the conductor would start and end on the same type of charge, which isn't possible.

 

[pixel]

That is not the reason
The reason is that on humid days things don't hold a charge so easily as the humid air allows it to leak away
rather that build up.

That's true, but it's also true that for a given amount of charge on your hand the ability to ionize the air depends on the humidity level. Water is a polar molecule and the water molecules line up in a manner to reduce the potential difference between your hand and the metal.

 

[Dale]

I can't visualize why there wouldn't be an electric field inwards.

What is there to visualize? If there were an electric field inside there would be a current. Is that clear?

In an electrostatic situation there is no current, by definition.

 

[Sho Kano]

Add enough charges, so many that we can stop thinking about them as discrete objects and more like a continuous circle of charge, and you'll find that the field lines no longer go inward at all.

The field lines going inward only cancel out each other in the situation of a circle right?

 

[Sho Kano]

What is there to visualize? If there were an electric field inside there would be a current. Is that clear?

In an electrostatic situation there is no current, by definition.

Yes, it's clear. By no electric field, you mean the net electric field inside is zero right?

 

[davenn]

That's true, but it's also true that for a given amount of charge on your hand the ability to ionize the air depends on the humidity level.

you are missing the point I made
the humidity stops a charge building up, period. The moisture in the air makes the air a better conductor and allows
the charge to continuously drain awayDave

 

[Dale]

Yes, it's clear. By no electric field, you mean the net electric field inside is zero right?

Yes.

 

[Sho Kano]

I have another question regarding electric fields:
If we place a conducting balloon in an electric field, and the outside field is pointing to the left. There will be an induced polarity with the negative charges clumping to the left side. It is observed that the balloon moves to the left. WHY does it do this? The superposition of the two fields shown is 0? - unless the outside field is "stronger" than the inside because the balloon "runs out" of electrons.

Here is another problem:
The inside of the balloon contains no charges. So if we apply Gauss's law, there will be no E field. But in this situation, there is clearly an E field.

 

[my2cts]

The static field inside the metallic balloon is zero everywhere.
The sum of the external field and the fields caused by the redistribution of charges exactly cancels.
This is true for any volume enclosed by a conductor, regardless of its shape.
See https://en.wikipedia.org/wiki/Faraday_cage .

 

[Drakkith]

The inside of the balloon contains no charges. So if we apply Gauss's law, there will be no E field. But in this situation, there is clearly an E field.

No there isn't. You haven't drawn it correctly.

Draw an insulating balloon in an external e-field. You'll find that the e-field passes right through the balloon since the charges making up the balloon cannot move about.

If I suddenly turn the balloon's material into a conducting material, the charges are now free to move about and will redistribute themselves. They will move until their own electric field exactly cancels out the external e-field inside the balloon.

The field lines going inward only cancel out each other in the situation of a circle right?

The easy, simple answer is yes. As with everything, it gets much more complicated when you get into the details.

If we place a conducting balloon in an electric field, and the outside field is pointing to the left. There will be an induced polarity with the negative charges clumping to the left side. It is observed that the balloon moves to the left. WHY does it do this? The superposition of the two fields shown is 0? - unless the outside field is "stronger" than the inside because the balloon "runs out" of electrons.

I'm not so sure the balloon will move if the external field is uniform. In such a field the force on every charged particle should be equal. If the field is stronger on the left side than the right side however, then you'd have a situation where the negative charges have moved into the region where the field is stronger. The attractive force would be stronger on these charges than the repulsive force on the positive charges on the right side of the balloon.

 

[Sho Kano]

They will move until their own electric field exactly cancels out the external e-field inside the balloon.

Yes, the negative charges in the conducting material of the balloon will move towards the field lines, and leave the positives behind. This leads to a field inside the conductor (from positive to negative charges) that cancels out the outside field. So how did I draw it incorrectly?

 

[Drakkith]

Yes, the negative charges in the conducting material of the balloon will move towards the field lines, and leave the positives behind. This leads to a field inside the conductor (from positive to negative charges) that cancels out the outside field. So how did I draw it incorrectly?

There isn't a field inside the balloon so you shouldn't have any arrows inside the balloon. The field set up by the balloon's charges opposes the external field and cancels it, so if you're drawing the overall electric field then you would not have any arrows inside the balloon. Note that Gauss's law is about the net flux through a closed surface. Inside the balloon the flux through a closed surface is always zero, regardless of whether or not you have an e-field passing through, since there are no charges enclosed. Any field lines passing through one side of this surface simply come out the other, so their overall contribution is zero.

 

[Guest812]

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?

If excess charges are placed on a very long thick conducting flat plate, the excess charges will redistribute themselves until their sideways (tangent to the surface) forces are equal and opposite to each other. When the all the charges on a conducting plate are redistributed so all their sideways (tangential) forces are equal and opposite, the charges will stop moving sideways (tangentially). When the charges stop moving they will equally force each other sideways (tangentially), and perpendicularly outwards from the conductor. The surface charge's sideways E-fields will cancel each other, which explains why the surface charges stop moving sideways. However the surface charge's E-fields will point perpendicularly in both opposite directions from the conductor's surface. This means on each side of the thick conductor there will be perpendicular directed E-field lines. This means inside the conductor there will be two oppositely directed E-fields, one E-field from each side's surface charges, which combine to a net of zero E-field. Outside the conductor, the E-fields from both side's surfaces charges add in the same direction.

In other words, a charge's E-field lines always continue forever, even inside a conductor; however, the charges distribute themselves on the surface of a conductor so the E-fields from all the charges cancel each other inside the conductor.

Example of positive excess charge E-fields on thick conductor:

(combined E-fields from both surfaces) <=== E | (+) E ---> <--- E (+) | E===> (combined E-fields from both surfaces)
(combined E-fields from both surfaces) <=== E | (+) E ---> <--- E (+) | E===> (combined E-fields from both surfaces)
(combined E-fields from both surfaces) <=== E | (+) E ---> <--- E (+) | E===> (combined E-fields from both surfaces)

simplifies to:

(combined E-fields from both surfaces) <=== E | (+) (zero net E-field) (+) | E ===> (combined E-fields from both surfaces)
(combined E-fields from both surfaces) <=== E | (+) (zero net E-field) (+) | E===> (combined E-fields from both surfaces)
(combined E-fields from both surfaces) <=== E | (+) (zero net E-field) (+) | E===> (combined E-fields from both surfaces)

where:
| = conductor surface
(+) = excess positive charge
<--- = E-field from one surface charge side
<=== = combined E-fields from both surface charge sides

 

[Sho Kano]

If excess charges are placed on a very long thick conducting flat plate, the excess charges will redistribute themselves until their sideways (tangent to the surface) forces are equal and opposite to each other. When the all the charges on a conducting plate are redistributed so all their sideways (tangential) forces are equal and opposite, the charges will stop moving sideways (tangentially). When the charges stop moving they will equally force each other sideways (tangentially), and perpendicularly outwards from the conductor. The surface charge's sideways E-fields will cancel each other, which explains why the surface charges stop moving sideways. However the surface charge's E-fields will point perpendicularly in both opposite directions from the conductor's surface. This means on each side of the thick conductor there will be perpendicular directed E-field lines. This means inside the conductor there will be two oppositely directed E-fields, one E-field from each side's surface charges, which combine to a net of zero E-field. Outside the conductor, the E-fields from both side's surfaces charges add in the same direction.

In other words, a charge's E-field lines always continue forever, even inside a conductor; however, the charges distribute themselves on the surface of a conductor so the E-fields from all the charges cancel each other inside the conductor.

Example of positive excess charge E-fields on thick conductor:

(combined E-fields from both surfaces) <=== E | (+) E ---> <--- E (+) | E===> (combined E-fields from both surfaces)
(combined E-fields from both surfaces) <=== E | (+) E ---> <--- E (+) | E===> (combined E-fields from both surfaces)
(combined E-fields from both surfaces) <=== E | (+) E ---> <--- E (+) | E===> (combined E-fields from both surfaces)

simplifies to:

(combined E-fields from both surfaces) <=== E | (+) (zero net E-field) (+) | E ===> (combined E-fields from both surfaces)
(combined E-fields from both surfaces) <=== E | (+) (zero net E-field) (+) | E===> (combined E-fields from both surfaces)
(combined E-fields from both surfaces) <=== E | (+) (zero net E-field) (+) | E===> (combined E-fields from both surfaces)

where:
| = conductor surface
(+) = excess positive charge
<--- = E-field from one surface charge side
<=== = combined E-fields from both surface charge sides

That was very clear! However inside the conductor, only the exact middle will experience no net field, so towards either side, there would be a net field pointing to the center?

 

[Dale]

so towards either side, there would be a net field pointing to the center?

If that were true then there would be a current.

 

[Sho Kano]

If that were true then there would be a current.

That's very non-intuitive because if a charged particle was closer to one surface (inside the conductor) than the other, I feel like it would experience a net force.

 

[Guest812]

That was very clear! However inside the conductor, only the exact middle will experience no net field, so towards either side, there would be a net field pointing to the center?

No, for a very long flat conducting plate (regardless of how thick the plate is), the E-fields will be constant no matter how far perpendicular away you from the surface, regardless of whether you're measuring from the charged surface's left or right side.

I recommend you study how to calculate the E-field from an infinitely thin, infinitely long, flat plane of surface charge. It can be calculated from Coulomb's law and calculus, but it is easier to calculate using Guass' Law. When you do, you'll find the E-field is constantly directed perpendularily outward equally from both sides of the infinitely thin charged surface, and is:

E = (excess charge) / (2 * permitivity )

Note in the above equation the E-field is always constant regardless of how far away from the surface.

Then after you've calculated the E-field from an infinitely thin, infinitely long, flat plane of surface charge, in a diagram simply add another infinitely thin, infinitely long, flat plane of surface charge parallel with the first plane, and your final diagram should look like the crude diagram I tried to type in my above preceding post. Therefore:

Outside the thick conductor having excess charge, equally distributed on both outer surfaces, the E-field at each surface will be (directed away from the conducting surfaces):

E = (excess charge) / (permitivity)

Note outside the thick conductor (with surface charges on both sides of the conductor, the E-field is twice as strong as it would have been on one side of an infinitely thin surface charge. This is because outside the thick conductor, both infinitely thin surface charges contribute E-fields in the same directions, and therefore add constructively.

Inside the thick conductor, both infinitely thin surface charges contribute E-fields in opposite directions, and therefore add destructively.

 

[Guest812]

That's very non-intuitive because if a charged particle was closer to one surface (inside the conductor) than the other, I feel like it would experience a net force.

The excess charges do experience a force. The charges force each other away from each other, like a ballon's wall is forced outward when you add more "gas charge" inside the balloon. The main difference between how an excess "gas charge" inside a balloon behaves and an excess electrical charge inside a conductor behaves, is the elastic balloon can stretch; whereas the metallic conductor is crystalline, meaning the conductor's atoms have rigidly fixed positions relative to each other. If the excess charge on a conductor's surface becomes too strong, the resulting E-field will be so strong that it can force the extremely excess charge to leave the conductor's surface and accelerate into the air. That is how a spark is generated.

 

[Dale]

The charges do experience a force.

Not inside the conductor.

 

[Sho Kano]

No, for a very long flat conducting plate (regardless of how thick the plate is), the E-fields will be constant no matter how far perpendicular away you from the surface, regardless of whether you're measuring from the surface's left or right side.

I recommend you study how to calculate the E-field from an infinitely thin, infinitely long, flat plane of surface charge. It can be calculated from Coulomb's law and calculus, but it is easier to calculate using Guass' Law. When you do, you'll find the E-field is constantly directed perpendularily outward equally from both sides of the infinitely thin charged surface, and is:

E = (charge) / (2 * permitivity )

Note in the above equation the E-field is always constant regardless of how far away from the surface.

Then after you've calculated the E-field from an infinitely thin, infinitely long, flat plane of surface charge, in a diagram simply add another infinitely thin, infinitely long, flat plane of surface charge parallel with the first plane, and your diagram should look like the crude diagram I tried to type in my above preceding post. Therefore:

Outside the thick conductor:

E = (charge) / (permitivity)

Note outside the thick conductor (with surface charges on both sides of the conductor, the E-field is twice as strong as it would have been on one side of an infinitely thin surface charge. This is because outside the thick conductor, both infinitely thin surface charges contribute E-fields in the same directions, and therefore add constructively.

Inside the thick conductor, both infinitely thin surface charges contribute E-fields in opposite directions, and therefore add destructively.

Yes, I understand this already; in post 42 and 44 I was referring to a circular conductor. (a sphere)

 

[Dale]

That's very non-intuitive because if a charged particle was closer to one surface (inside the conductor) than the other, I feel like it would experience a net force.

I understand, but that is why you need to think systematically and logically first. Your intuition will often be wrong and you need experience before you will be able to improve it.

Remember, in a conductor $J=\sigma E$ by definition. So if you have an electrostatic situation then you must have $E=0$

 

[Sho Kano]

I understand, but that is why you need to think systematically and logically first. Your intuition will often be wrong and you need experience before you will be able to improve it.

Remember, in a conductor $J=\sigma E$ by definition. So if you have an electrostatic situation then you must have $E=0$

Yes by definition there has to be no electric field, else paired electrons below the surface will start flowing.
Maybe thinking about this problem in the way of electric fields is misleading, though there must be a way to resolve this using them (?)