Surface Charge Density, Polarization

AI Thread Summary
The discussion centers on calculating the surface charge density at the upper surface of water when all molecular dipoles are aligned downward. The user attempts to apply the equation for bound surface charge density, assuming it is equivalent to the overall surface charge density. They derive a value of approximately -0.205 C/m² using the dipole moment of water and relevant density and molecular mass data. There is uncertainty regarding the assumptions made about the direction of polarization and the relationship between bound and overall surface charge density. The conversation emphasizes the importance of clearly stating assumptions in the calculations.
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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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