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- Thread starter arun_mid
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James R

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The positive and negative charges on opposite plates attract each other.

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mukundpa

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lightgrav

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and the second has the identical number of excess negative charges,

all the E-field (lines) will start on the positives and end on the negatives.

There won't be any E-field in the material, so no potential difference.

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Let the charge on the inner surface of one of the plates be q. Then, construct a guassian cylinder, with one face inside one of the plates, and the other face inside the other plate. Since there's no electric field inside a conductor in electrostatics, the flux through this cylinder must equal zero. So the net charge inside the guassian cylinder is zero. Hence, the charge on the inner sides of the plates must be equal and opposite in sign.

Hence, the other plate's inner surface has charge -q. Now the remaining charge on the first plate is Q-q, and on the second, -Q+q. However:

Consider any point within the first plate. The electric field at the point is zero, because it is inside a conductor. But the fields due to the two inner plates cancel out; they are equal in magnitude and opposite in direction. Hence, the fields due to the outer surfaces must also cancel. But the point lies IN BETWEEN these outer surfaces, hence the charges on the two outer surfaces are equal in magnitude and sign!

So, Q-q=-Q+q.

Hence, q=Q.

Therefore, charge on outer surface = 0.

However, I was thinking of something else, a case where the above equations can't be used as such. In that case, no mathematical equations will easily tell you that the outer surface has no charge... I'll post it as soon as I remember.

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arun_mid said:

Let the charge on the inner surface of one of the plates be q. Then, construct a guassian cylinder, with one face inside one of the plates, and the other face inside the other plate. Since there's no electric field inside a conductor in electrostatics, the flux through this cylinder must equal zero. So the net charge inside the guassian cylinder is zero. Hence, the charge on the inner sides of the plates must be equal and opposite in sign.

Hence, the other plate's inner surface has charge -q. Now the remaining charge on the first plate is Q-q, and on the second, -Q+q. However:

Consider any point within the first plate. The electric field at the point is zero, because it is inside a conductor. But the fields due to the two inner plates cancel out; they are equal in magnitude and opposite in direction. Hence, the fields due to the outer surfaces must also cancel. But the point lies IN BETWEEN these outer surfaces, hence the charges on the two outer surfaces are equal in magnitude and sign!

So, Q-q=-Q+q.

Hence, q=Q.

Therefore, charge on outer surface = 0.

However, I was thinking of something else, a case where the above equations can't be used as such. In that case, no mathematical equations will easily tell you that the outer surface has no charge... I'll post it as soon as I remember.

This is incorrect.

As a simple conceptual experiment, make your capacitor out of spheres

instead of plates and deform them continuously into flat plates. The spheres

have charge all over them (nonuniformly distributed.)

The charge distribution is related to the electric field outside the wires

as well as the plates. There will be a small charge on the outside of the

plates as well as up the wires leading away from the capacitor. The charge

decreases with distance away from the plates as you move up the wire, but

is nowhere zero.

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Antiphon said:This is incorrect.

As a simple conceptual experiment, make your capacitor out of spheres

instead of plates and deform them continuously into flat plates. The spheres

have charge all over them (nonuniformly distributed.)

But can't the charges redistribute so that the charge on the outer surfaces becomes zero? What's mathematically wrong with my argument? As for nonuniform distribution, the inner plates may have slightly nonuniform distribution, however the distribution must be the same on both plates. Gauss law demands it.

Antiphon said:The charge distribution is related to the electric field outside the wires as well as the plates. There will be a small charge on the outside of the plates as well as up the wires leading away from the capacitor. The charge decreases with distance away from the plates as you move up the wire, but is nowhere zero.

Why should charge appear on the wire surface? Before electrostatic condition, the wire should have no net charge, although there's a constant flow of negative charge towards the positive terminal. When electrostatic condition is reached, why a change in it's nature? Or is it because the wires connected to the two plates form another capacitor? I ignored this effect because it shouldn't affect the charge on the plates.

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mukundpa

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Simply bring a test charge from infinity(very large distance) to the positive plate of a charged capacitor along the central line out side the plate, if there is no field outside, the work done should be zero. Then what is the potential of the positive plate.

Or

Think you are charging the capacitor along the outside central line then if there is no charge out side the capacitor the workdone will be zero. From where the capacitor gets energy?

|+ - |

|+ - |

|+ - |-------

|+ - |

|+ - |

Or

Think you are charging the capacitor along the outside central line then if there is no charge out side the capacitor the workdone will be zero. From where the capacitor gets energy?

|+ - |

|+ - |

|+ - |-------

|+ - |

|+ - |

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arun_mid said:But can't the charges redistribute so that the charge on the outer surfaces becomes zero? What's mathematically wrong with my argument?

No. In order for this to happen, the electric field just outside the capacitor

would have to be identical with the electric field of a wire without a capacitor.

The only situation where this is true is an identically zero field, or V=Q=0.

As for nonuniform distribution, the inner plates may have slightly nonuniform distribution, however the distribution must be the same on both plates. Gauss law demands it.

Yes, but it is not demanded by Guass's law. It is demanded by the symmetry

of the problem. If one of the plates is shaped differently, there will be different charge distributions, and this is not addresed by Guass's law.

Why should charge appear on the wire surface? Before electrostatic condition, the wire should have no net charge, although there's a constant flow of negative charge towards the positive terminal. When electrostatic condition is reached, why a change in it's nature? Or is it because the wires connected to the two plates form another capacitor? I ignored this effect because it shouldn't affect the charge on the plates.

I don't understand your question here.

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mukundpa said:Simply bring a test charge from infinity(very large distance) to the positive plate of a charged capacitor along the central line out side the plate, if there is no field outside, the work done should be zero. Then what is the potential of the positive plate.

Or

Think you are charging the capacitor along the outside central line then if there is no charge out side the capacitor the workdone will be zero. From where the capacitor gets energy?

|+ - |

|+ - |

|+ - |-------

|+ - |

|+ - |

I never said that the field outside is zero. I really have no idea about that...I said that a point INSIDE one of the plates is very close to the inner sides of the two plates. If the two plates have area A, and a charge q is stored in the capacitor, then field due to positive plate(This is due only to the inner side, which therefore can be considered as a charged sheet, not as a thick metal plate) for all points nearby is

[tex]\\ E \ = \ \frac{q}{2A\epsilon_0} [/tex]

Whereas the other plate, being also close by, produces a field

[tex]\\ E \ = \ - \frac{q}{2A\epsilon_0} [/tex]

Hence the field equals zero there. You can compare it to a dipole. At large distances there will be an appreciable field. Nearby the field effectively equals zero.

When a capacitor charges, the electrons reach the negative plate, displacing an equal number of electrons from the positive plate. Hence the work done is the same as that done when you push an electron from the positive plate to the negative plate. The field outside plays no role. See any derivation for energy of a parallel plate capacitor.

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arun_mid said:I never said that the field outside is zero. I really have no idea about that...

This is exactly why there are charges all over the capacitor and all

the way up the wires to infinity in both directions- because the field

is not zero anywhere.

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Antiphon said:This is exactly why there are charges all over the capacitor and all

the way up the wires to infinity in both directions- because the field

is not zero anywhere.

Well, how does the field not being zero anywhere lead to charge on outer surface of the plates being non-zero? I mean, if there were charge on the outer plates of the capacitor, then what about what I had said earlier?

What's wrong there?Let the charge on the inner surface of one of the plates be q. Then, construct a guassian cylinder, with one face inside one of the plates, and the other face inside the other plate. Since there's no electric field inside a conductor in electrostatics, the flux through this cylinder must equal zero. So the net charge inside the guassian cylinder is zero. Hence, the charge on the inner sides of the plates must be equal and opposite in sign.

Hence, the other plate's inner surface has charge -q. Now the remaining charge on the first plate is Q-q, and on the second, -Q+q. However:

Consider any point within the first plate. The electric field at the point is zero, because it is inside a conductor. But the fields due to the two inner plates cancel out; they are equal in magnitude and opposite in direction. Hence, the fields due to the outer surfaces must also cancel. But the point lies IN BETWEEN these outer surfaces, hence the charges on the two outer surfaces are equal in magnitude and sign!

So, Q-q=-Q+q.

Hence, q=Q.

Therefore, charge on outer surface = 0.

please explain using plates of equal area for simplicity.

And as for the charge distribution stuff, I could certainly make a guassian cylinder with a differential cross sectional area, that had one face inside each plate, and claim that since the flux was zero, the differential charge on that differential area [tex]\\dq \ = \ \sigma dA [/tex]

of one plate would correspond to -dq on the other and hence the charge distribution would be equal and opposite... if the other plate had a different area then I would say: superimpose one plate over the other. the charge will be distributed only on the common area of the two plates.

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arun_mid said:Well, how does the field not being zero anywhere lead to charge on outer surface of the plates being non-zero? I mean, if there were charge on the outer plates of the capacitor, then what about what I had said earlier?

The tangential component of the elecric field must be zero on the conductor.

If you were to cut the wires off your capacitor, the elctric field would have

a tangential component in the space where the surface of the wires were.

When the wires are there, this tangential component is compensated by a charge

distribution along the wires- all the way to inifinity.

What's wrong there?

please explain using plates of equal area for simplicity.

And as for the charge distribution stuff, I could certainly make a guassian cylinder with a differential cross sectional area, that had one face inside each plate, and claim that since the flux was zero, the differential charge on that differential area [tex]\\dq \ = \ \sigma dA [/tex]

of one plate would correspond to -dq on the other and hence the charge distribution would be equal and opposite... if the other plate had a different area then I would say: superimpose one plate over the other. the charge will be distributed only on the common area of the two plates.

Sorry, I have to step out. I will come back to this shortly.

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Alternately, if there WEREN'T wires, would my logic work?

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arun_mid said:Alternately, if there WEREN'T wires, would my logic work?

I don't think so.

Let's connect two widely seperated metal spheres by wires and a battery.

The spheres are made of a neutral metal so initially each has a balance of

+ and - charges.

S ------ (-B+) ------ S

When we connect the battery, it pushes electrons onto the

negative-side sphere and pulls electrons off the positive-side sphere.

Then we disconnect the wires and battery and take them away.

Each sphere will have charges distributed almost uniformly over their surfaces,

but there will be a small extra concentration of electrons on the - sphere

where it is closer to the distant + sphere, and there will be a slightly

reduced concentration of electrons on the + sphere where it faces the

- sphere. Nevertheless, the charges will be almost completely uniformly

distributed.

If you now deform the plates into cubes, the same situation holds except

that the corners of the boxes will have extra charges bunching up there.

But all six sides of the cube will have excess + or - charge.

Now flatten the boxes into thin plates. Much of the charge will gather

on the edges and points but the front and back will have almost equal

charges for each plate.

Finally, as you bring the plates closer together, the opposite charges

will tend to concentrate on the sides of the plates facing each other

If what you say were true, can you tell me at what seperation distance

the plates will be when the entire charge on each plate moves to only

the one side?

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There has to be some distance between the plates at which the charge is nearly totally on the inner side.

Remember that a point inside the plate has to have no net field.

If there were equal charges on the back and front side of the plate, then fields due to the back and front sides would cancel out at any point inside the plate, but the other plate is oppositely charged, and it would cause a net electric field at that point, wouldn't it?

So, I think you're right--the charge increases slowly on the inner sides as they draw closer...we came to the same conclusion for small distances! But I think there will be a lot of difference between the charges on inner and outer sides at a distance of about, say, 0.1mm! for plates about 5cm^2 in area...

Anyway, I think I got your point here, Antiphon...thanks for the help. But can you explain the situation when there are wires attached again? I couldn't understand what you meant there...

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if the plates were inifinite in size or (in the limit) as the separation became zero.

As for the attached wires, it is difficult without pictures but I will try.

First, you have a capacitor without wires like this.

(p) ................................. +||-

From the symmetry of the problem, we know that the electric field at

the test point "p" is a vector which points toward the left side of the screen.

Now let us connect wires to the capacitor like so.

----(p)-----------------------+||------------------------------------

The electric field paralell to the wire at (p) must now be zero because it is

tangential to the (perfect) conductor. This means that the presence of

the wire has changed the electric field there. It does this by distributing

charges along its length which produce the modified field.

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mukundpa

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consider the diagram in the file attached. The plates are very close and large.

plate A is given + ve and B is given equal -ve charge. There will be a Potential difference say V. It means that the work done to bring a unit charge from a point on surface 3 to surface 2 along the arrow is V. Now think, what will be the work done in bringing a unit charge from a point on the surface 4 to the surface 1, through a very large curved path outside the capacitor. Is it zero?

If yes then how it possible because A and B are conducting and the inner surfaces 2 and 3 are having non zero potential difference V.

If no then which field is creating this P.D.?

Actually plate a is the combination of two capacitors in parallel, because charge is stored on both surfaces

1. The charge on inner surface is due to formation on a parallel plate capacitor between the two surfaces 2 and 3 of the plates and

2. The charge on the outer surface 1 of A .

(Think, a single spherical conductor is having capacity = 4 *Pi*epsilon zero then why a conductor of the shape of a single plate will not have capacity and why not each surface? )

The Net capacity is the sum of the two but the contribution of the surface is very very small so we neglect this as compared with the capacitance of the inner capacitor but it is not zero. . . . .**negligible but not zero...**

plate A is given + ve and B is given equal -ve charge. There will be a Potential difference say V. It means that the work done to bring a unit charge from a point on surface 3 to surface 2 along the arrow is V. Now think, what will be the work done in bringing a unit charge from a point on the surface 4 to the surface 1, through a very large curved path outside the capacitor. Is it zero?

If yes then how it possible because A and B are conducting and the inner surfaces 2 and 3 are having non zero potential difference V.

If no then which field is creating this P.D.?

Actually plate a is the combination of two capacitors in parallel, because charge is stored on both surfaces

1. The charge on inner surface is due to formation on a parallel plate capacitor between the two surfaces 2 and 3 of the plates and

2. The charge on the outer surface 1 of A .

(Think, a single spherical conductor is having capacity = 4 *Pi*epsilon zero then why a conductor of the shape of a single plate will not have capacity and why not each surface? )

The Net capacity is the sum of the two but the contribution of the surface is very very small so we neglect this as compared with the capacitance of the inner capacitor but it is not zero. . . . .

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reilly

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You will find this problem worked out in some freshman physics books. There's an old saying in physics: don't use a piledriver to crack an egg. If you want a bit more of a challenge, replace the conducting plates by dielectric plates.If you want an even harder problem, let the plates oscillate.

Regards,

Reilly Atkinson

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Anyway, I think I understood what Antiphon presented...thank you, Antiphon. It now strikes me that we can consider the wires to be a infinite series of capacitors...

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