Charge on a parallel plate capacitor


by tomwilliam2
Tags: capacitor, charge, parallel, plate
tomwilliam2
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#1
Sep24-12, 05:56 PM
P: 64
1. The problem statement, all variables and given/known data
If a parallel plate capacitor is charged with charge Q and -Q on its plates, respectively, and then isolated, then we place a dielectric into the space between the plates, does the charge on the plates decrease, or stay the same.


2. Relevant equations
U=(QV)/2


3. The attempt at a solution
I saw Walter Lewin's lecture in which he says that the surface charge density is changed after you put the dielectric in place (sigma_0 becomes sigma_0 - sigma_induced) and that made sense to me. The induced polarisation of the dielectric material displaces some bound charge onto the surface of the plates, creating an induced electric field to counter the existing field.
However, in a tutorial tonight one of my tutors argued that the total charge Q stays the same after the dielectric is put in place, the capacitance goes up and the voltage goes down.
Are these two explanations compatible? How can the areal surface charge density decrease but the charge on the plates remains the same?
Thanks in advance
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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Pranav-Arora
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#2
Sep24-12, 10:08 PM
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In the Walter lewin's lecture, was the capacitor disconnected from the battery?
tomwilliam2
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#3
Sep25-12, 01:59 AM
P: 64
Yes, he specifically states that the capacitor is charged up and then the battery is disconnected and taken away.

ehild
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#4
Sep25-12, 05:07 AM
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Charge on a parallel plate capacitor


Can then the charge of one plate change? The plate is isolated...

ehild
tomwilliam2
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#5
Sep25-12, 05:25 AM
P: 64
I would have said no, the charge must remain constant. I was surprised then to see Walter Lewin's lecture (MIT Electromagnetism, polarisation and capacitors, the second instalment, around minute 2 and a half). He argues that the charge density on the plate (sigma_free) is changed when the dielectric is placed between the plates due to the polarisation of the material. The new charge density is now sigma_free - sigma_induced. He argues that the induced charge is responsible for the electric field which acts oppositely to the existing field due to the potential difference, and which reduces the net electric field.
It seems to make sense, but then so does my tutor's explanation of how charge remains constant (in which he mainly uses algebraic methods with known formulae for capacitance, etc.)
Any ideas?

Could it be that the free charge Qf remains constant, while the surface charge density has to be altered with the addition of a bound surface charge?
ehild
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#6
Sep25-12, 11:20 AM
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The charge on the capacitor plates can not change if the plates are isolated. But the dielectric becomes polarized by the electric field between the plates. Dipole chains form and the ends of these chains appear as surface charge of opposite sign as the charge on the metal plate. This surface charge belongs to the dielectric, but interact with the charges on the plate and neutralize them so they do not contribute to the electric field any more. Therefore the electric field decreases if a dielectric slab is placed between the plates of the capacitor, and also the voltage decreases.
Walter Levin and your tutor says the same, although they might use different terms in the explanation. The electric field decreases, as the number of free charges decreases. Or the electric field decreases as the induced surface charge causes opposite electric field.

ehild
Attached Thumbnails
dipolechains.JPG  
tomwilliam2
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#7
Sep25-12, 04:22 PM
P: 64
Thanks, that makes perfect sense.


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