How does the presence of a dielectric change the max V of a capacitor?

[Leo Liu]

Homework Statement

Title is the question. It is a conceptual one and is not for marks.

Relevant Equations

:)

My guess is that while the voltage between the two plates is lower when a dielectric is present, the maximum voltage that the capacitor can hold will actually increase because the maximum strength electric field generated by the charges on the two plates will be higher due to the opposing electric field created by the dielectric. If we assume that the magnitude of the E field at which an electrical breakdown can occur does not change when the dielectric is added (although I am not sure whether this assumption is valid), then the maximum of the voltage goes up due to ##\Delta V_{max}=E_{max}d##.

What's the correct answer to this question?

 

[ergospherical]

If the relative permittivity is denoted by ##\epsilon##, then one may introduce an auxiliary field ##\mathbf{D} = \epsilon_0 \epsilon \mathbf{E}## which satisfies the equation ##\nabla \cdot \mathbf{D} = \rho_{\mathrm{f}}##, where ##\rho_{\mathrm{f}}## is the free charge density (i.e. not including bound charges in the dielectric). What does this version of Gauss' law imply for the electric field strength in your problem?

 

[Leo Liu]

...

that the field strength will be smaller?

 

[ergospherical]

Yep, by a factor of ##\epsilon##. By extension the voltage is reduced, and the capacitance increased, by that same factor.

 

[Steve4Physics]

It is true that, for a given charge, a dielectric will reduce the voltage between the plates.

But the question appears to be about maximum (breakdown) voltages. It looks like you are being asked to compare the maximum sustainable voltages for:
a) a capacitor with a vacuum between its plates and
b) the same capacitor but with a material dielectric
irrespective of how much charge is stored.

The maximum (breakdown) voltage with a vacuum is determined by field-emission – when the electric field is so high that electrons are pulled from a capacitor plate’s surface.

The maximum breakdown voltage with a material dielectric is determined by the material’s dielectric strength. Breakdown starts when the field inside the dielectric is strong enough to pull outer electrons from the atoms.

Edit.
It is worth noting that capacitors for very high voltage applications always often use a vacuum. But note there are materials with a higher breakdown strength than a vacuum. E.g. take a look at this: https://en.wikipedia.org/wiki/Vacuum_variable_capacitor

 

[Leo Liu]

It is true that, for a given charge, a dielectric will reduce the voltage between the plates.

But the question appears to be about maximum (breakdown) voltages. It looks like you are being asked to compare the maximum sustainable voltages for:
a) a capacitor with a vacuum between its plates and
b) the same capacitor but with a material dielectric
irrespective of how much charge is stored.

The maximum (breakdown) voltage with a vacuum is determined by field-emission – when the electric field is so high that electrons are pulled from a capacitor plate’s surface.

The maximum breakdown voltage with a material dielectric is determined by the material’s dielectric strength. Breakdown starts when the field inside the dielectric is strong enough to pull outer electrons from the atoms.

edi - typo's

It is worth noting that capacitors for very high voltage applications always use a vacuum.

Thanks a lot Steve. This is the answer I am looking for!

 

[Steve4Physics]

Thanks a lot Steve. This is the answer I am looking for!

I have updated my answer - see edit at end of Post #5.