Surface Charge Density, Polarization

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SUMMARY

The discussion focuses on calculating the surface charge density at the upper surface of water when all molecular dipoles are aligned downward. The electric dipole moment for water is given as \( p = 6.13 \times 10^{-30} \, C \cdot m \). The surface charge density is derived using the formula \( \sigma_b = P \cdot \hat{n} \), resulting in \( \sigma_b = -0.205 \, \frac{C}{m^2} \). The user seeks confirmation on the assumptions regarding bound surface charge density and the direction of the polarization vector.

PREREQUISITES
  • Understanding of electric dipole moments and their significance in molecular physics.
  • Familiarity with the concepts of polarization and bound charge density.
  • Knowledge of basic chemistry, including molar mass and density calculations.
  • Proficiency in vector notation and directional analysis in physics.
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  • Study the relationship between electric dipole moment and polarization in materials.
  • Learn about the differences between bound and free charge density in dielectric materials.
  • Explore the implications of molecular alignment on macroscopic electrical properties.
  • Investigate the role of dipole moments in other polar molecules beyond water.
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Homework Statement



The electric dipole moment for the water molecule equals $$ p = 6.13 × 10−30 C · m $$ Suppose that in the glass of water all molecular dipoles could be made to point down. Calculate the resulting surface charge density at the upper water surface

Homework Equations


[/B]
## P = ## dipole moment per unit volume

## \sigma_b = P \cdot \hat{n} ##

## D_{water} = 1,000,000 [\frac{g}{m^3}] ##

## M_{water} = 18.02 [\frac{g}{mol}]##

## N_A = 6.022 *10^{23} [\frac{1}{mole}] ##

The Attempt at a Solution



I'm not really sure how to go about this one, I've seen other similar problems online, and here, but the book seems kind of vague to me on this topic, for some reason.

I wasn't sure if "bound" surface charge density would be the same as surface charge density in this situation, so I just went ahead and assumed it is for this problem :

$$ \sigma_b = P \cdot \hat{n} = (\frac {mp} {V_u}) \cdot \hat{n} $$, where ## m = ## number of molecules being polarized, ## V_u = ## unit volume, and ## \hat{n} ## is of negative orientation due to the surface being the opposite direction of the polarization of the molecules?

So, now I need to figure out ##m## and ## V_u##, volume is mass divided by density, so I assume I need the atomic mass, and density of water...

$$ \sigma_b = - (\frac {D_{water}} {M_{water}}) N_a p = -(3.3*10^28[\frac {1}{m^3}])(6.13*10^-30[C \cdot m]) = -.205 [\frac {C}{m^2}]$$

I just need some confirmation on my work in this problem, I'm really not sure about assuming ##\sigma_b ## is the same as ##\sigma## and also, in terms of the direction of ## \hat{n} ##. It just seemed weird also including all this old chemistry knowledge in a physics textbook that never once mentions anything about moles, so I'm not sure if all this was needed.
 
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The reasoning seems fairly sound to me ... you are modelling the mater molecule as a pair of charges: when the dipole points down, one of the charges is on the top. Then you need the surface density of water dipoles and the size of the uppermost charge. That the units check out should be a clue you did OK.
When you write it out be clear about the assumptions you made.
 
Simon Bridge said:
The reasoning seems fairly sound to me ... you are modelling the mater molecule as a pair of charges: when the dipole points down, one of the charges is on the top. Then you need the surface density of water dipoles and the size of the uppermost charge. That the units check out should be a clue you did OK.
When you write it out be clear about the assumptions you made.

Okay, thanks. I just wasnt too sure about the negative direction of the polarizability, and also that the bound surface density was the same as the overal surface density.
 

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