Flux Density in Capacitor

Why are you assuming that V is constant?

There are two formulae for C:

C = k A/d and C = Q/V

So: kA/d = Q/V

if you change k by introducing a dielectric with different k the equality can only balance if something else changes.

A and d are fixed. Q can only change of there is a conductive path between the plates. That only leaves V.

Also, think energy.

E=0.5CV^2.

For this equality to balance, if C increases due to the dielectric, V has to decrease.

Actually, the author Robert Boylestead of Introductory Circuit Analysis gives a what if situation,

"with or without the dielectric - if the potential across the plates is kept constant and the distance between the plates is fixed, the net electric field withing the plates must remain the same, as determined by the equation Electric Field = V/d."

Boylestead states that since the electric field of the dielectric (due to dipoles) opposes the electric field of the plates, more charge must be deposited onto the plates from source power to maintain a constant resultant electric field of the capacitor. Boylestead gives equation:

Electric Field = Q / ( k * A)

So it sounds to me that Q is changing.

Boylestead gives equation:

k = D / Electric Field

Therefore if k goes up, the flux density would have to go up. This is where I get stumbled. Isn't flux density and electric field pretty much same thing?

If you want to have real fun with this, consider what happens when you charge an air gapped cap with a fixed voltage, disconnect the source, then insert a dielectric into the gap. Of course the voltage goes down, since the charge on the plates is not changed, but the C has gotten bigger. What about conservation of energy, one might ask. In the air case, the energy is 1/2 C V^2. If the material inserted increases C by 2, the new voltage is V/2 by definition, so the new energy is

1/2 (2*C)V^2/4, 2C because the value of the cap has increased by 2, and V^2/4 because the voltage had gone down by a factor of 2.

But the new energy is half the old energy. It's a little bit of a head scratcher for engineers in training.

AJW

Yes, and if you go 88 miles per hour you can go back in time.

It's not quite that simple. You have to have 1.21 jigga-watts of power too! ...or a Mr. Fusion. ;-)

Yep, that is another option, hence my comment re the conductive path:

Either way

kA/d = Q/V

has to balance.

If the potential is held constant then Q must change, which only happens if charge can move ie there is a conductive path. No conductive path, then V must change.

Actually, this equation says that k goes up if Electric Field goes down.

Brian

? that doesn't conserve energy the way you've written it:

E, energy, constant but C doubling then V has to decrease by a factor of

1/root(2)- so "new" V^2 is equal to "old" (V^2)./2- then the energy equation balances.

Not if you remember the 1/root(2) ;-)

Look again at the V/2 bit. It should be V/(2^0.5) HTH

Chimera

I suggest you read Brian's post again then say sorry for trying to feed back his information in a different way.

Chimera

Don't tell me this is DVS is Airy under a new name?

Chimera

Brian, you have the patience of job.

Chimera

Oh, I am sorry, I thought the post said flux density capacitor.