And I'm still smiling

The gate is not one single isolated conductor.

and my maths tells me that 1/2CV^2 = 0 since v = 0 on an isolated conductor.

..............................

I though we were talking about the energy it takes to place one electron on a capacitor, not an isolated conductor (which really makes no sense to me).

 

I though we were talking about the energy it takes to place one electron on a capacitor, not an isolated conductor (which really makes no sense to me)

I thought that my post53 was as clear as I could make it.

You cannot put a single (or any odd number) charge on a connected capacitor, only pairs of charges since as soon as you insert one charge onto one plate the charges on the other will adjust to balance in accordance with conservation of charge.

When you have two charges you have potential energy since each charge is in the electric field of the other (though conventionally we consider one as generating all the field and the other as a 'test' charge).
A single charge generates a field, but no potential energy, but can only be found in a truly unconnected capacitor.

So every time to add a pair of charges to a capacitor you add to the electric potential energy and thereby increase the voltage across the plates.

 

So many have already put their oar in here that I hesitate to add to the confusion, but you will not resolve without abandoning some basic misconceptions.

1) We do not 'charge' inductors. No charge is separated or stored in an inductor.

2) The creation/depletion of a magnetic field is dependent on the rate of change of current. At the beginning of charging a capacitor the rate of change is very large, but this reduces as time goes on so near the end of the charging the rate of change of current is very small. This is why basic charge discharge curves for capacitors are studied in elementary physics.

3) The work required to place the first charge on and uncharged capacitor is precisely zero. When you try to add a second charge you have to overcome the repulsion provided by the first one so have so do a small amount of work. Each additional charge added requires more work than the previous one as it faces greater repulsion. The total work is the sum of each of these individual quantities. When we go from placing charges one at a time to a continuous current we move from a sum to an integral.

From:
http://www.allaboutcircuits.com/vol_1/chpt_15/1.html

"When the current through an inductor is increased, it drops a voltage opposing the direction of electron flow, acting as a power load. In this condition the inductor is said to be charging, because there is an increasing amount of energy being stored in its magnetic field."

"Conversely, when the current through the inductor is decreased, it drops a voltage aiding the direction of electron flow, acting as a power source. In this condition the inductor is said to be discharging, because its store of energy is decreasing as it releases energy from its magnetic field to the rest of the circuit."

This is just semantics.

From:
http://en.wikipedia.org/wiki/Work_(physics)
"In physics, a force is said to do work if, when acting on a body, there is a displacement of the point of application in the direction of the force."

Or, work is a force acting through a distance.

When that first charge is placed on the capacitor, is work required to repel the same polarity charge off the opposite plate? Was there a force acting through a distance that displaced the opposite charge? If the first charge required no work to be placed on the plate, then is no work require to displace the opposite charge?

The theory that a capacitor stores (1/2)QV does not have special conditions for the voltage
value. The voltage could be very near zero or very large.

 

We have been through this before.

A capacitor has two distinct conductors.
What happens to the second conductor if you introduce a single charge (say an electron) on to the first?

That depends upon what the second conductor is connected to.

1) If the second conductor is isolated from the rest of the universe, nothing.
No work is done.
The capacitor remains 'uncharged'
No voltage difference is induced between the plates.

2) If the second conductor is connected to an electron sink

An electron is displaced from the second conductor to the sink
~Work is done
A voltage between the two plates is now established
The capacitor is now charged.

But once again work is only done when the second charge is introduced, or in this case, leaves, the system.

Now work is done?

3) The work required to place the first charge on and uncharged capacitor is precisely zero. When you try to add a second charge you have to overcome the repulsion provided by the first one so have so do a small amount of work. Each additional charge added requires more work than the previous one as it faces greater repulsion. The total work is the sum of each of these individual quantities. When we go from placing charges one at a time to a continuous current we move from a sum to an integral.

Where you refering to a capacitor with a single plate here? Zero work to place the first charge on an uncharged single plate capacitor?

 

ele
Where you refering to a capacitor with a single plate here? Zero work to place the first charge on an uncharged single plate capacitor?

Ther is no such thing as a single plate capacitor.

studiot
A capacitor has two distinct conductors.

Let us work backwards through the rest of your questions and a more complete and thouough definition of work should become apparent.
You Wiki definition is a definition of mechanical work, and should also include a statement about direction.

There is also electrical work, that is the work of electrical forces on electrical charge, which is what we are talking about here.

The electrical work required to place two charges Q1 and Q2 a distance a apart, in an otherwise empty universe is

\(W = \frac{{Q1Q2}}{{4\pi \varepsilon a}}\)

If there is only one charge, Q2 does not exist so Q2 = 0 so W = 0.
So no electrical work is done.

The theory that a capacitor stores (1/2)QV does not have special conditions for the voltage
value. The voltage could be very near zero or very large.

This is perfectly true.
And in an empty universe with one uncharged capacitor V= 0
Adding one single charge to one of the aforementioned terminals does not change V as Q2=0 in the above equation.
So 1/2QV = 0

"When the current through an inductor is increased, it drops a voltage opposing the direction of electron flow, acting as a power load. In this condition the inductor is said to be charging, because there is an increasing amount of energy being stored in its magnetic field."

"Conversely, when the current through the inductor is decreased, it drops a voltage aiding the direction of electron flow, acting as a power source. In this condition the inductor is said to be discharging, because its store of energy is decreasing as it releases energy from its magnetic field to the rest of the circuit."

This is just semantics.

Paradoxically magnetic forces are unable to do electrical work.
This is because the magnetic force is always at right angles to the displacement it causes to the charge.

This is why I say you cannot charge an inductor.

Thank you for bringing that part of the Ebook to our attention.
I think it is wrong use of terminology.

I Think of it like this

If I charge up a capacitor and remove tha capacitor from the source and put it in my pocket and take it somewhere else, the capacitor will still be charged

Just like if I charge up my cell battery or my credit card or whatever.

But if I connect an inductor to a source and than disconnect it, and take it aswy, it will contain zero charge.

 

Ther is no such thing as a single plate capacitor.


Let us work backwards through the rest of your questions and a more complete and thouough definition of work should become apparent.
You Wiki definition is a definition of mechanical work, and should also include a statement about direction.

There is also electrical work, that is the work of electrical forces on electrical charge, which is what we are talking about here.

The electrical work required to place two charges Q1 and Q2 a distance a apart, in an otherwise empty universe is

\(W = \frac{{Q1Q2}}{{4\pi \varepsilon a}}\)

If there is only one charge, Q2 does not exist so Q2 = 0 so W = 0.
So no electrical work is done.


This is perfectly true.
And in an empty universe with one uncharged capacitor V= 0
Adding one single charge to one of the aforementioned terminals does not change V as Q2=0 in the above equation.
So 1/2QV = 0




Paradoxically magnetic forces are unable to do electrical work.
This is because the magnetic force is always at right angles to the displacement it causes to the charge.

This is why I say you cannot charge an inductor.

Thank you for bringing that part of the Ebook to our attention.
I think it is wrong use of terminology.

I Think of it like this

If I charge up a capacitor and remove tha capacitor from the source and put it in my pocket and take it somewhere else, the capacitor will still be charged

Just like if I charge up my cell battery or my credit card or whatever.

But if I connect an inductor to a source and than disconnect it, and take it aswy, it will contain zero charge.

Sounds good to me studiot. I think I should let this go, but always more questions. I think your saying there is no Q2 on the plate Q1 is being placed on, so no Q2 and no work. Does Q2 not exist on the opposite plate and require work to displace?

 

Before we start there is no Qanything on the plate we are placing Q1 on.
There is also no Qanything on the second plate either.

Since there are no charges there is no electric field.

When we introduce Q1 there is only one charge, which generates an electric field.
This field contains no energy by itself.
Q1 has not moved through any electrical field since there isn't one until it arrives.

If the second plate we have not added any charge to is connected to a suitable charge source/sink a balancing charge Q2 is rapidly drawn to the second plate.
In moving to the second plate Q2 moves through the field of Q1 and experiences a force according to Coulomb's law.
So work is done, depending upon the signs, on or by Q2 in making this move.

So Q2, the second charge to arrive in the system is the first charge where any work in done.
Once Q2 is in the system it would take energy to move it away again. This is the potential energy it has in the capacitor, which we conventionally attribute to the field in the capacitor dielectric.