# Change of relative permittivity of liquid and solid water

Dear Physics Forum Users
Commonly, the relative permittivity of liquid water is reported to be $\epsilon_r = 78.0\epsilon_0$, $\epsilon_0$ being the dielectric constant of the vacuum.
For ice (solid water), $\epsilon_r = 4 \epsilon_0$ (heard it in a talk once).

Is it correct to interpret the different values in this way that ice is around 20 times less polarizable than liquid water? Which intuitively would make sense.

Thanks for comments.

## Answers and Replies

Simon Bridge
Science Advisor
Homework Helper
Is it correct to interpret the different values in this way that ice is around 20 times less polarizable than liquid water?
How would you measure "polarizableness"?

$\epsilon_0$ would be the permittivity of free space - "the dielectric constant of the vacuum" would be 1. This is important in relation to above since you may want to consider a substance more polarizable in proportion to the magnitude of the polarization density for a given applied electric field ... the polarization density of a vacuum is zero.

How would you measure "polarizableness"?

$\epsilon_0$ would be the permittivity of free space - "the dielectric constant of the vacuum" would be 1. This is important in relation to above since you may want to consider a substance more polarizable in proportion to the magnitude of the polarization density for a given applied electric field ... the polarization density of a vacuum is zero.

Thanks for your comment, however I dont understand it:
What is the difference between free space and vacuum formally?

Do you agree to the statement that polarization arises from the response of a polarizable material to an electric field, i.e. the dipole moments of the material (built-in or induced) align with the field which creates a new, opposing field, which partially cancels the applied field within the material, albeit not completely (except for within metals).

How then are the different relative permittivities $\epsilon_r = 4.0$ for ice and $\epsilon_r = 78$ for water to be interpreted?

nasu
Gold Member
Is it correct to interpret the different values in this way that ice is around 20 times less polarizable than liquid water? Which intuitively would make sense.

Thanks for comments.

Before you do this you should make sure that the two values are measured at the same frequency. For your case, I suspect that they are not.
The value of about 80 is the relative permittivity of water in DC or static fields. It decreases significantly at high frequencies.
The value of about 4 (for ice) may be the one measured in the microwave range, like in this paper (for example):
http://eis.bris.ac.uk/~glhmm/pdfs/WestJEEG07.pdf

Simon Bridge
Science Advisor
Homework Helper
Thanks for your comment, however I dont understand it:
What is the difference between free space and vacuum formally?
http://en.wikipedia.org/wiki/Vacuum

"free space" is less ambiguous.
In classical electromagnetism, "the vacuum of free space" is distinct from just any old vacuum.

However that wasn't the distinction I was going for.

Do you agree to the statement that polarization arises from the response of a polarizable material to an electric field
Yes.

Consider:
If I have a material with relative permittivity 20, does it make sense to say that this material is 20x more polarizable than a vacuum (relative permittivity of 1)?

Last edited:
Consider:
If I have a material with relative permittivity 20, does it make sense to say that this material is 20x more polarizable than a vacuum (relative permittivity of 1)?

you're saying it does not make sense, because the vacuum is not polarizable..

why do i need to measure the dielectric constant of ice depending on frequency? can I not just insert a block of ice between two electrodes and apply a potential to one side?

From $C = \epsilon_r \epsilon_0 \frac{A}{d}$ and $C=Q/U$, can I not get $\epsilon_r$, simply once I know $U$ and $Q$?

Given such an experiment, I place a piece of ice and a volume of water in the capacitor and measure the relative constant of dielectricity for both, classically, what do the two different values tell me about the material?