# Dielectrics and Capacitance

1. Feb 21, 2013

### CAF123

1. The problem statement, all variables and given/known data
A parallel plate capacitor with plate area A and plate separation d=5mm is connected to a 120 Volt power supply and allowed to fully charge.

A) Calculate the capacitance, stored charge, electric field midway between the plates and
potential energy stored.

The capacitor is disconnected from the power supply and a sheet of glass of thickness b = 2mm with dielectric constant κ = 5 in place between midway between plates.

B)Calculate the electric field inside the dielectric and in the gaps between the dielectric
and the capacitor plates, hence calculate the voltage across the plates, and thus the new capacitance.

C) Calculate the new stored energy and determine whether work was done or given out
as the glass plate was inserted.

2. Relevant equations
Capacitance, dielectric eqns

3. The attempt at a solution

A)capacitance = Q/V, Q = σA = (Eε)A, For the electric field midway, I can assume this is a constant if I take the length of the plates >> distance between them. So simply E = V/d.

B)In the gaps, E field in A) divided by kappa. I am not sure about the E field actually inside the dielectric. (I suppose it depends on the nature of the dielectric - if it was a conductor it would be zero I think). Since it is given that we have glass, I am not sure. I know it induces an E-field but I think this applies outside the dielectric.
New voltage: E down by kappa, d same => V down by kappa. (120/kappa)V

C) I notice that $U_{free} > U_{ind}$ (U before dielectric greater than afterwards). So the capacitor has done work against the insertion of the dielectric. Physical reasoning: Capacitor no longer connected to power supply, so all charge is fixed. When the dilectric comes in , the charge densities on the plates will create temporary dipoles on the insulator. I am not so sure why this implies work has to be done against the dielectric, since surely there would be an attraction as a result of the dipole.

Many thanks.

2. Feb 22, 2013

### ehild

Have you learnt about the vector of electric displacement? It is usually denoted by D, and DE.

Because of the planar geometry, both D and E are normal to the capacitor plates.

At the interface of two media, the normal component of D is the same at both sides of the interface. So D(air)=D(glass) → ε0E(air)=εglassE(glass).

Also because of the planar geometry, the equipotential surfaces are parallel with the capacitor plates. You can insert a thin metal plate along an equipotential surface. So you can handle the capacitor with the inserted glass slab as three capacitors connected in series, and having the same charge as the original one.

ehild

Last edited: Feb 22, 2013
3. Feb 22, 2013

### CAF123

Hi ehild,
Sorry, no we haven't covered the electric displacement vector. What is εglass?

I can't think of another way to compute the E-field inside the dielectric.

4. Feb 22, 2013

### ehild

Well, then the E field is E0/kappa inside the dielectric if E is the field in the air gap.
It is easier for you then to use the three-capacitors in series method.
ε is the permittivity of a dielectric. κ is the ratio ε/ε0, or the relative permittivity.

ehild

Last edited: Feb 22, 2013
5. Feb 22, 2013

### CAF123

I think the results I gave in the OP are only valid if the dielectric fills the whole space between the two parallel plates.

I wrote previously that in the gaps between the dielectric and capacitor the E field will decrease by kappa. This is not true since a quick check with Gauss Law confirms. However I do have one question about this (see attachment).

As you said, the E-field inside the dielectric is E_ind = E_free/kappa. I also have another question about this (see attachment).

The new voltage across the plates would then by E_free(3mm) + E_ind(2mm), correct?

Why would the capacitor be doing work against the dielectric?

Many thanks.

#### Attached Files:

• ###### Capacitance.png
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6. Feb 22, 2013

### ehild

The method to calculate the new voltage is correct.
As for the green Gaussian surface: The electric field is the same on both the upper and lower faces, but the normal vectors of the faces are opposite. Remember the definition of flux.

ehild

7. Feb 22, 2013

### CAF123

Do you mean the ochre Gaussian surface? On the top of the bottom plate, the normal points upwards (and this is antiparallel to the downwards E field). I see.

8. Feb 22, 2013

### CAF123

Do you have any thoughts/comments on my other questions?

9. Feb 22, 2013

### ehild

The energy of the capacitor will decrease. So the capacitor does work on the dielectric, by polarizing it and attract inside. As for your questions in the picture, I do not follow you. Draw one Gaussian and ask one question at one time, please.

ehild

10. Feb 22, 2013

### CAF123

Yes, sorry I worried that my picture may have been too cluttered.

See my first question attached.

Many thanks.

#### Attached Files:

• ###### Capacitance2.png
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11. Feb 22, 2013

### ehild

When you put a closed surface like in the figure into a homogeneous electric field the flux on the top face is opposite to the flux on the bottom. Although the net flux on the whole Gaussian surface is zero, the electric field is not zero.

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