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Surface charge density on the outer surface of a plate in a parallel plate capacitor

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dharm0us
#1
May27-09, 01:41 PM
P: 7
Let me know if the following reasoning is correct :

If the surface charge density
on the internal surface of one of the
plates in a parallel plate capacitor
is

Sigma,

which becomes
Sigma/K,

when a dielectric
material is inserted between the plates with
dielectric constant K,
so the charge density on the outer
surface of the same plate will be

Sigma - Sigma/K
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Bob S
#2
May27-09, 04:04 PM
P: 4,663
If you have a parallel plate capacitor with charge +/-Q on the inside of each plate, the voltage between the two plates is (d=separation, A = area)
V=Q/C = Qd/e0A
If you now stick a dielectric with relative dielectric constant k into the gap, the voltage becomes
V=Q/C = Qd/k e0A
The unchanged charge remains on the inside of both plates. The stored energy is now
W = (1/2) Q V = Q2d/k e0A
So the stored energy is reduced when you inserted the dielectric. Where did it go?
If there is any unbalanced charge on the outer surface of either plate, it will remain there. Gauss law will tell you that the E field lines from charges on the outside of the plates are still there, and terminate at infinity.
dharm0us
#3
May27-09, 11:33 PM
P: 7
Hi Bob S,
Let me rephrase my question :
When a dielectric is inserted, some charge is induced
on the inside of the plate, so the opposite charge of
equal amount will be induced on the outside of the plate.

Is it correct?

Actually, I am making a question, to be handed out to
students, based on the above assumption, so I wanted
to cross check if the above reasoning is correct indeed.

Phrak
#4
May27-09, 11:48 PM
P: 4,512
Surface charge density on the outer surface of a plate in a parallel plate capacitor

Charge-up a plate capacitor. Let things settle until there is no current. The electric potential on both sides of a plate will be equal to each other.

All the charge is on the surface of the conductor, so we talk about the surface charge density. Although the electric potential is equal throughout the volume of the plate, the charge density is not.

The charge will be redistributed upon the surface to equalize the electric potential.
dharm0us
#5
May28-09, 12:58 AM
P: 7
Hi Phrak,
So, is the reasoning in my original post is correct?
Phrak
#6
May30-09, 12:35 AM
P: 4,512
Quote Quote by dharm0us View Post
Hi Phrak,
So, is the reasoning in my original post is correct?
You have proposed that the surface charge density is the same on both surfaces-the one facing the oppositely charged plate and the outer plate.

What should draw electrons to the outward facing surface? You need to specify the environment or no answer can be given.

I presume you mean that there are no external charges present, in which case the charge density on the external side is zero.
dharm0us
#7
May30-09, 01:53 AM
P: 7
Hi Phrak,

What I have proposed is this :
Initially there is no charge on the outer surface of the plate,
and the surface charge density on the inner surface is
Sigma.

When the dielectric is inserted,
the charge density on the inner surface
is = Sigma / K, where K is the dielectric constant of
the material.

So, rest of the charge moves on the outer surface.
Hence the surface charge density on the outer surface
becomes :

Sigma - Sigma/K
Born2bwire
#8
May30-09, 05:50 AM
Sci Advisor
PF Gold
Born2bwire's Avatar
P: 1,776
No, the charge density is just sigma/k. The rest of the charges will migrate back between the connected leads to the opposite plate. In essence, the stored energy does not change as the loss in power stored in the electric field between the plates due to the lessening of the charges is made up by energy stored in the polarization of the dielectric. This assumes that you still have a DC potential hooked up between the plates.

If you have the parallel plate capacitor charged, at steady state, then disconnect the leads, and then move a dielectric in between the two plates, the charge on the plates is still sigma, the charge has nowhere to go (barring dielectric breakdown, etc). The difference here is that moving the dielectric between the plates will require extra work, this extra work is stored in the polarization of the dielectric. So since you have to add energy into the system, there is no violation of energy conservation.
DaleSpam
#9
May30-09, 10:28 AM
Mentor
P: 17,202
Quote Quote by dharm0us View Post
Actually, I am making a question, to be handed out to
students, based on the above assumption, so I wanted
to cross check if the above reasoning is correct indeed.
Is it really fair to your students to ask them a question that you are not confident enough yourself to answer it without assistance? I think you are getting into a potentially bad situation here.

That said, I don't think there is enough information. Are the two capacitor plates charged with some voltage source that remains connected? Are they charged and then disconnected? Etc.
Bob S
#10
May30-09, 12:04 PM
P: 4,663
If there is a charge +/-sigma on the inside of both plates, and the plates are NOT connected to a power supply, when a dielectric with constant k is inserted into the gap, the charge sigma will remain the same, and the voltage between the plates will drop by a factor k.
If there is a charge +/-sigma on the inside of both plates, and the plates ARE connected to a power supply, when a dielectric with constant k is inserted into the gap, the charge sigma will drop by a factor k, and the voltage between the plates will be unchanged.
If there is no charge on the OUTSIDE of the capacitor at the beginning, meaning Gauss' Law surface integral outside the capacitor =0, there are no field lines to (and charges at) infinity,
there will be no outside charges after the dielectric is inserted.
In either case, the total energy stored in the capacitor is reduced because the stored energy W is
W = (1/2) Q V = (1/2) C V2 = (1/2)Q2/C = (1/2)Q2d/k e0A.


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