B Question about understanding conductors for EM course

[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 (?)

 

[Dale]

I wouldn't say it is misleading. It is just unfamiliar to you and so you have not yet built up a correct intuition.

 

[Guest812]

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

If the sphere diameter is infinitely large, near the infinitely large sphere's surface, the sphere's E-fields will behave as if the sphere's surface was flat.

A tiny sphere, can treated as an infinitely large sphere, if you measure extremely close to the sphere's surface. If you know how the E-fields behave extremely close to the sphere's surface, then by using symmetry and logic, you can probably determine how they behave further away on both sides of the surface.

Or in other words, a sphere (regardless of diameter) having excess surface charge, will have radially directed E-fields, directed both outward away from the center and inward toward the center all around the sphere. The inwardly directed E-fields will totally cancel each other everywhere within the sphere, but add to the outwardly directed E-fields on the opposite side.

 

[Sho Kano]

but add to the outwardly directed E-fields on the opposite side.

I didn't know about this, but it makes sense because a charged particle would experience a force on the other side due to THAT surface charge.

The inwardly directed E-fields will totally cancel out each other everywhere within the sphere

Yes, so if I place a charged particle at a radius (lets say the radius of the sphere is 1 meter) 0.75 meters from the center of the sphere, it would experience no force. So there is a crazy thing going on that the surface charges don't contribute at all inside the conductor.

 

[Guest812]

...

Yes, so if I place a charged particle at a radius (lets say the radius of the sphere is 1 meter) 0.75 meters from the center of the sphere, it would experience no force. So there is a crazy thing going on that the surface charges don't contribute at all inside the conductor.

Yes.

In my above posts, I diagrammatically explained it using a thick flat conducting plate with equal surface charge on both opposite sides analogy, then explained how a conducting sphere of infinite diameter could be treated as a conducting flat plate, and how even the tiniest conducting sphere when "viewed" extremely close to its surface can be treated as a sphere of infinite diameter. No matter what shape the surface is, when viewed extremely close to its surface, that infinitesimally small surface area will appear as an infinitesimally small flat plate surface area.

The best "no E-field anywhere inside a conducting sphere" explanation involves diagrams and calculus. The easiest mathematical approach I've seen involves Guass' Law, and the "Solid Angle" concept. But as I explained above, before attempting this more mathematically challenging spherical problem, it would probably be wiser (easier) to mathematically solve first the infinitely thin single charged plane, and then solve the two parallel infinitely thin surfaces charges separated by a thick long flat conducting plate. All the decent college calculus based Physics books explain those flat plane calculations, and the better Physics books will mathematically derive why the E-field is zero everywhere inside a conducting sphere, as well as why the E-field is zero anywhere inside any shaped conductor.

 

[Sho Kano]

Or for other words, a sphere (regardless of diameter) having excess surface charge, will have radially directed E-fields, directed both outward away from the center and inward toward the center all around the sphere.

Right, a flat conducting plane will have a constant field as long as you are close to it. So now, since any sphere can be treated as a plane (close up), and planes have a constant field, it wouldn't matter if you were .75 from the center. Is this what you're getting at? Again, this is very non-intuitive.

 

[Guest812]

Right, a flat conducting plane will have a constant field as long as you are close to it. So now, since any sphere can be treated as a plane (close up), and planes have a constant field, it wouldn't matter if you were .75 from the center. Is this what you're getting at? Again, this is very non-intuitive.

The text I change to red is incorrect, in the context you used it in. The correct and more accurate phrasing is:

"an infinitely long flat conducting plane will have a constant E-field, regardless of how far away from the plane the E-field is measured.

Note an infinitely long flat conducting plane will look exactly the same, regardless of how close you are to it. If you're an inch away from an infinitely long surface, take a photo, then magnify the photo so an inch distance looks like a mile, the plane will still look infinitely long, but now you look like you're a mile away from the plane.

Like others and I have been repeatedly trying to explain to you, before you think about the E-field inside a sphere, it would be easier if you first understand the E-field on both sides of a long infinitely thin conducting surface charge, then think about the E-field on both sides of a long thick conducting surface with surface charges on both sides. These examples are standard beginner college level Physics E&M problems. If you can't solve these beginner type problems, you'll probably be less able to solve more mathematically advance problems like calculating the E-field inside a conducting sphere.

 

[Sho Kano]

The text I change to red is incorrect, in the context you used it in. The correct phrasing is:

"a long flat conducting plane will have a constant field, regardless of how away from it you are.

Only in the case of an infinitely large plane. There are none, and I can't see how that logic can be applied to a non-infinite, limited sphere of radius r.

before you think about the E-field inside a sphere, it would be easier if you first understand the E-field on both sides of a long infinitely thin conducting surface charge,

Like I said, I already understand that. The field is given by \frac { \sigma }{ 2{ \varepsilon }_{ 0 } }

then think about the E-field on both sides of a long thick conducting surface with surface charges on both sides.

Putting the two parallel to each other (assume negative charges on both), just superposition the fields. On top of the plate is just \frac { \sigma }{ { \varepsilon }_{ 0 } }In the middle is 0. Below is again, \frac { \sigma }{ { \varepsilon }_{ 0 } }.

 

[Guest812]

Only in the case of an infinitely large plane. There are none.

Like I said, I already understand that as I already completed my EM course. The field is given by \frac { \sigma }{ 2{ \varepsilon }_{ 0 } }

Putting the two parallel to each other (assume negative charges on both), just superposition the fields. On top of the plate is just \frac { \sigma }{ { \varepsilon }_{ 0 } }In the middle is 0. Below is again, \frac { \sigma }{ { \varepsilon }_{ 0 } }.

Good.

Then from there, the only way I know from the top of my head to intuitively view what happens inside and outside a conducting sphere, is visualize how the E-field behaves when it is measured infinitely close to both sides of the sphere's surface, where infinitely close to a sphere's surface means so close that the sphere appears to be flat. Like the way the Earth appears when you stand on Earth, you're so close to the Earth's round surface that for all practical purposes the Earth's round surface appears flat to you.

Chapter 4 of this free downloadable E&M text from MIT explains how to calculate and visualize E-fields inside and outside conductors very well: http://ocw.mit.edu/courses/physics/...ng-2007/class-activities/chapte4gauss_law.pdf

Also the free Video Lecture from Yale's Physics II 201 course taught by professor Shankar has excellent explanations and mathematical derivations: http://oyc.yale.edu/physics/phys-201/lecture-4

 

[Sho Kano]

visualize how the E-field behaves when it is measured the infinitely close to both sides of the sphere's surface, where infinitely close to a sphere's surface means so close that the sphere appears to be flat.

Yes, like in post 56, the field is constant.

Like you way you feel when you stand on Earth, you're so close to the Earth's round surface that for all practical purposes it appears flat to you.

Like how gravitational field can be approximated to g? What if I go closer to the center, does it still stay constant?

 

[Guest812]

Yes, like in post 56, the field is constant.

Like how gravitational field can be approximated to g? What if I go closer to the center, does it still stay constant?

NO !

A sphere only appears as a flat surface when you're extremely close to the sphere's surface. The further away from the sphere you are, the more the sphere will appear as a point charge.

In other words, when viewed extremely close to the conducting sphere's surface, the E-field lines emanate perpendicular from both externally and internally from a sphere's surface, which when viewed extremely close to the sphere's surface the E-field lines appear to emanate parallel to each other; however, perpendicular to the spheres surface when viewed far away from the surface is actually radial: meaning externally from the sphere the further away from the sphere the weaker the E-field (i.e. externally the radial E-fields aren't constant, instead they weaken as 1/R^2), and internally the closer towards center the stronger the E-field; however for each internal E-field emanating from one side of the sphere there will be an exact opposite inward radially directed E-field emanating from the opposite side of the sphere. Since every point of the sphere generates an inward E-field exactly opposite of the opposite side of the sphere, every E-field is exactly canceled inside the sphere, the same way they did inside an infinitely long thick flat conducting plate.

I have a (bad) habit of editing (almost all my posts) after posting them. I think the best explanations will be found in the last two comments at the end of my earlier post #59 of this thread.