# Purcell 10.15: Water Dipoles

1. Dec 2, 2012

### ResonantW

1. The problem statement, all variables and given/known data
Imagine that all the molecular dipoles in a cup of water could be made to point down. Calculate the magnitude of the resulting surface charge density at the upper surface of the water, and express it in electrons per square centimeter.

2. Relevant equations

3. The attempt at a solution
Can I just find the number of water molecules per square centimeter of area, knowing the approximate size of an H2O molecules, and then say that since all the dipoles point down we will have 2 electrons for every such molecule along the surface?

2. Dec 2, 2012

### TSny

I think you should approach it from the ideas developed in chapter 10. The text gives you the permanent dipole moment, p, of a water molecule. It also explains how to go from dipole moment per molecule to the polarization P. From P you can get the surface charge density.

3. Dec 2, 2012

### ResonantW

Got it! Why wouldn't the first method work?

Also, do I know the volume density of water molecules, because that would be what Purcell calls N, correct?

4. Dec 2, 2012

### TSny

I don't think you can assume two electrons per molecule. The effective surface charge per surface molecule would depend on the amount of polarization of each molecule, p .
Right. You'll need to determine N, the number water molecules per cm3. See if you can calculate it from the mass density of water and the molar mass of water.

5. Dec 4, 2012

### ResonantW

I get that P= 6e22 esu/cm^2. Is that equivalent to the surface charge density?

I can also express this as 1.3e32 electrons per square centimeter.

6. Dec 4, 2012

### TSny

I think you overlooked the caption for Fig. 10.14 which states that the dipole moments given in the figure are in units of $10^{-18}$ esu-cm. Otherwise, looks good.