Confusion about Electric Permittivity

Question

I have learned about electric permittivity in two contexts:

1. How much a medium reduces the strength of the electric field (Like if I have a charged ball inside a vat of oil, the oil will polarize and create its own electric field in the opposite direction that reduces the strength of the net field)
2. How high capacitance is in a parallel plate capacitor ($C$=electric permittivity of the dielectric in between the plates * area of the plates / distance between the plates)

However, I feel like these two definitions are contradictory. In the capacitance context, I learned that the less polarizable something is, the better an insulator it is, and therefore it has a greater electric permittivity because electrons have a harder time jumping between plates so the plates can be more charged per volt (At least that's how it was explained to me - it might have been a simplification). So in this definition, less polarizable substances have higher permittivities.

But in the context about electric field, it would make sense to me that more polarizable substances have higher permittivities. My logic is that the more polarizable something can be, the greater an opposing electric field it can create when polarized (For example, Argon gas would diminish the field strength greater than Helium gas because each molecule has more electrons so they have a greater degree of polarization - like in London Dispersion Forces).

My understanding leads to one definition saying that less polarizable substances have higher permittivity and the other definition saying the opposite. Where is my mistake?

Answer 1

My understanding leads to one definition saying that less polarizable substances have higher permittivity and the other definition saying the opposite. Where is my mistake?

I'm having trouble following your logic, but the greater the electrical permittivity, the lower the electric field strength, and the greater the capacitance. The greater the capacitance the easier it is to move electrons from one plate to the other per volt via an external circuit (electrons don't jump between plates).

The electric field between a parallel plate capacitor is given by

$$E=\frac{V}{d}$$

Reducing the electric field strength due to greater electrical permittivity for a given plate separation reduces the voltage across the plates. The voltage across the plates is the work required, per unit charge, to move the charge from one plate to the other.

So if the electric field is reduced, less work is required per unit charge to move the charge from one plate to the other. It is easier to move the electrons from one plate to the other, not harder.

Hope this helps.

Answer 2

I recommend that you try to forget the contents of your paragraph that starts 'However...'. The dielectrics that you put between the plates of a capacitor should all be insulators, whether they are easy or hard to polarise. [Polarisation is a separation of charges within molecules, not movement of charge through the dielectric.] And electrons don't jump between the plates – unless the capacitor is being mis-used.

Suppose we charge a capacitor and then disconnect the battery. If we now place an insulator between the plates it will be polarised and produce its own electric field opposing that due to the charges ($±Q$) on the plates themselves. So the resultant electric field strength between the plates will drop, and therefore the pd between them. So the capacitance ($Q/V$) will increase. The more easily polarisable the material the greater its permittivity, the lower $V$ and the greater $C$. There is no contradiction.