Capacitance and charge confusion

[Terocamo]

It was told many times by my teacher that a disconnected metal plate and an earthed metal plate, forming a capacitor, will have the same charge regardless their separation.
I understand how the charge of the disconnected metal play remains unchanged, but I don't really understand how the earthed plate has it's charge unchanged when it is moved.
Let me illustrate withe an example. If you put the earthed plate A at a distance of 1cm from the disconnected plate B, which possesses constant positive charge.
When plate A move to B, say reducing the distance to 0.5cm, plate A will experience a greater potential due to B. However, it is earthed, so it must have zero potential, and it will draw electrons from the ground to cancel out the extra potential.
And isn't plate A gaining charge then?

 

[Antiphon]

This is approximately true for ideal capacitors (very big plate, very small separation).

It becomes untrue in the general case. I'll give an example at the end.

What's confusing you is that a potential change does not require the flow of charge relative to the conductors. When two unearthed plates are pulled apart, the potential difference goes up but the charges on the plates are the same. This is also true when one of the plates is earthed. All it does is change the reference point for zero voltage. This does not imply any flow of charge.

Where the statement begins to fall apart is where your intuition is starting to lead you. If you really set up actual plates like this then you don't have one capacitor, you have a system of coupled capacitors. The disconnected plate also has measurable capacitance to the walls and earth, not just to the earthed plate.

Here's the obvious scenario that mixes it all together. I have a pair of plates, one earthed and one not. There is a charge Q1 on that pair. Nearby I have a second set of plates again one of them earthed with a second charge Q2. Now what happens to the validity of the original statement if I swap the positions of the unearthed plates? Clearly the charges will flow through the Earth so that the earthed plate charges will each change from Q1 to Q2 and vice versa.

 

[Studiot]

By 'disconnected' I assume you mean isolated.

and it will draw electrons from the ground to cancel out the extra potential.

Why do you think this?
Electrons don't cancel potential, they add to or subtract from the charge.

If the isolated plate has charge Q and it remains isolated, there is no supply of charge to it, so the charge cannot change.

A charged plate, does not exert any electric force or potential force on a nearby neutral (uncharged) object however large the charge.
What it does do it induce an equal on opposite charge (-Q) on the facing surface of the grounded plate.

The usual electric field then exists between the two opposite plates, and we can determine the electric field from Coulombs law.

Since the inducing charge Q cannot change, the induced charge (-Q) cannot change either.

The more you separate opposite charges the greater their potential.
Conversely the closer you bring them the lower their potential.

Since capacitance is the ratio of charge to potential, for a fixed charge an decrease in potential means an increase in capacitance.

Q = CV

Does this help?

 

[Antiphon]

A charged plate, does not exert any electric force or potential force on a nearby neutral (uncharged) object however large the charge.
What it does do it induce an equal on opposite charge (-Q) on the facing surface of the grounded plate.

Actually this is not quite correct. There are net forces and torques induced on the net charge-neutral object. This is also true for dielectric bodies.

If you introduce an uncharged conductor into a field of a fixed set if charges, the field energy is reduced. This implies a force on the conductor as it is introduced.

 

[Studiot]

Actually this is not quite correct. There are net forces and torques induced on the net charge-neutral object. This is also true for dielectric bodies.

If you introduce an uncharged conductor into a field of a fixed set if charges, the field energy is reduced. This implies a force on the conductor as it is introduced.

I think we are meant to be talking introductory electrostatics here.

 

[Antiphon]

Never pass of erroneous information as introductory. It's doing a disservice to the student.

 

[Studiot]

Actually this is not quite correct. There are net forces and torques induced on the net charge-neutral object. This is also true for dielectric bodies.

Is this a denial of Coulombs law?

Perhaps you could provide a calculation showing the force and torque exerted (electrically) by a proton on a neutron at a distance of 1 mm .

I calculate both of these to be zero.

 

[Terocamo]

By 'disconnected' I assume you mean isolated.
Why do you think this?
Electrons don't cancel potential, they add to or subtract from the charge.

If the isolated plate has charge Q and it remains isolated, there is no supply of charge to it, so the charge cannot change.

A charged plate, does not exert any electric force or potential force on a nearby neutral (uncharged) object however large the charge.
What it does do it induce an equal on opposite charge (-Q) on the facing surface of the grounded plate.

The usual electric field then exists between the two opposite plates, and we can determine the electric field from Coulombs law.

Since the inducing charge Q cannot change, the induced charge (-Q) cannot change either.

The more you separate opposite charges the greater their potential.
Conversely the closer you bring them the lower their potential.

Since capacitance is the ratio of charge to potential, for a fixed charge an decrease in potential means an increase in capacitance.

Q = CV

Does this help?

Can you illustrate a bit more about the induced charge? Why can't the induced charge change when the distance between the inducing charge and the induced charge varies?

In fact, this is the point i can't get through i guess. You see, the influence of fields is usually stronger when closer, so why won't moving an earthed object closer to a charged object cause more charge be induced on that earthed object?

If it is true that when an object gains (or lose) charge when the aforesaid distance changes, how do we apply the equation Q=CV on the metal plates where their charge isn't the same?

maybe i can illustrate my confusion this way
[PLAIN]http://dl.dropbox.com/u/20735614/illustration.JPG
this is a diagram discribing what are setted up before plate A is moved,
the vertical lines are the equalpotential line due to the charged plate B,
and now plate A is experiencing a potential of +50 due to plate B, but it is earthed, so it must draw electrons to creat a potential of -50V there in order to remain 0V

after the plate is moved
[PLAIN]http://dl.dropbox.com/u/20735614/illustration2.JPG
you can see now the plate A is experiencing a potential of +100V due to plate B
but still it needs to maintain 0V, and because plate B is disconnected, the movement of plate A shouldn't alter the potential B creates, but it would definitely lower the potential of plate B because plate A contrubutes to cancel the potential B creates, and the only way to do so would be to drawn more electrons.

So which part did i think wrong?