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Goldenwind
Jan27-08, 08:52 PM
1. The problem statement, all variables and given/known data
The dipole moment of the water molecule H2O is 6.17x10^-30 Cm. Consider a water molecule located at the origin whose dipole moment p_vec points in the positive x direction. A chlorine ion (Cl-), of charge -1.60x10^-19 C, is located at x=3.00x10^-9 meters. Assume that this x value is much larger than the separation d between the charges in the dipole, so that the approximate expression for the electric field along the dipole axis can be used.

Find the magnitude of the electric force, ignoring the sign, that the water molecule exerts on the chlorine ion.

2. Relevant equations
E = p / (2pi*e0*x^3)
E = F/q
F = kq1q2/r^2
p = qr

e0 = 8.85*10^-12
q1 = -1.6*10^-19
x = 3*10^-9
p = 6.17*10^-30
k = 9*10^9


3. The attempt at a solution
We need to find the force. F is in two equations:
E = F/q
F = kq1q2/r^2

The first is the force of the electric field, not the force between two items, so we want the later equation.

We have k
We have q1
We have r (Assuming r = x)
We don't have q2

To find q2, we use one of the following:
p = qr
E = p / (2pi*e0*x^3)
E = F/q

The first is possible
The second we can't use as we don't have E
The third we can't use as we don't have E or F

So, p = q2*r
q2 = 6.17*10^-30 / 3*10^-9
q2 = 2.057*10^-21

Now to compute the force using:
k = 9*10^9
q1 = -1.6*10^-19
q2 = 2.057*10^-21
r = 3*10^-9

F = kq2q2/r^2
F = -3.2912*10^-13
Ignoring sign, 3.2912*10^-13

This answer is incorrect.
Where is my mistake?

Goldenwind
Jan27-08, 10:10 PM
Been working on this all week, and this is all I have. It's an online assignment which as due at midnight EST (Aka, in 50 minutes).

Could someone take a look before then? If not, it's cool.

Goldenwind
Jan27-08, 10:40 PM
Solved it.

lithium311
Feb27-08, 06:36 PM
Could you show how you solved this? TIA