How Does Aligning Water Dipoles Affect Surface Charge Density?

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SUMMARY

The discussion focuses on calculating the surface charge density resulting from aligning water dipoles in a cup of water. Participants emphasize the importance of understanding the dipole moment, polarization, and the volume density of water molecules. The effective surface charge density is derived from the polarization (P), which is calculated to be 6e22 esu/cm², equivalent to approximately 1.3e32 electrons per square centimeter. The conversation highlights the necessity of using the correct molecular parameters and equations from the relevant physics literature.

PREREQUISITES
  • Understanding of molecular dipoles and their alignment
  • Familiarity with the concepts of polarization (P) and surface charge density
  • Knowledge of water's mass density and molar mass for calculating molecular density
  • Proficiency in using equations related to dipole moments as outlined in physics texts
NEXT STEPS
  • Calculate the volume density of water molecules using mass density and molar mass
  • Explore the relationship between dipole moment and polarization in detail
  • Investigate the derivation of surface charge density from polarization
  • Review relevant chapters in Purcell's "Electricity and Magnetism" for deeper insights
USEFUL FOR

Students and educators in physics, particularly those studying electromagnetism and molecular interactions, as well as researchers interested in the properties of water at the molecular level.

ResonantW
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Homework Statement


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.


Homework Equations





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?
 
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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. :smile:
 
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?
 
ResonantW said:
Got it! Why wouldn't the first method work?

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

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.
 
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.
 
ResonantW said:
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.

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.
 

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