- #1
Destroxia
- 204
- 7
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.