# How Does Inserting a Dielectric Affect a Charged Capacitor?

• CAF123
In summary: The net flux determines the enclosed charge. In summary, the conversation covers the calculation of various parameters of a parallel plate capacitor, including capacitance, stored charge, electric field, and potential energy. It also discusses the insertion of a dielectric material between the plates and the resulting changes in the electric field, voltage, and stored energy. The concept of electric displacement and the use of Gaussian surfaces to calculate flux are also mentioned. The conversation ends with a question about the work done by the capacitor on the dielectric.
CAF123
Gold Member

## Homework Statement

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.

## Homework Equations

Capacitance, dielectric eqns

## 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.

Have you learned 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:
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.

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:
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.

#### Attachments

• Capacitance.png
23.6 KB · Views: 435
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

ehild said:
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

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.

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

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

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

See my first question attached.

Many thanks.

#### Attachments

• Capacitance2.png
12.2 KB · Views: 419
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.

#### Attachments

• greenbox.JPG
7.4 KB · Views: 383

## 1. What is the difference between a dielectric and an insulator?

A dielectric is a material that can store electrical energy in the form of an electric field. An insulator is a material that does not conduct electricity.

## 2. How does a dielectric affect the capacitance of a capacitor?

A dielectric increases the capacitance of a capacitor by reducing the electric field between the plates. This effectively increases the amount of charge that can be stored on the plates for a given voltage.

## 3. What is the role of dielectrics in electronic devices?

Dielectrics are used in electronic devices to increase the capacitance and energy storage capacity of capacitors. They are also used to insulate and protect sensitive components from electrical interference.

## 4. Can a dielectric be used in place of a conductor?

No, a dielectric cannot be used in place of a conductor because it does not allow the flow of electric current. Conductors are necessary for the functioning of electronic circuits.

## 5. How do you calculate the capacitance of a capacitor with a dielectric?

The capacitance of a capacitor with a dielectric can be calculated using the formula C = εA/d, where C is the capacitance, ε is the permittivity of the dielectric material, A is the area of the plates, and d is the distance between the plates. This formula takes into account the effect of the dielectric on the capacitance.

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