The dipole moment of the water molecule (H,O) is6.17 x 10-30 C.m. Consider a water molecule located at theorigin whose dipole moment p points in the positive x direction. Achlorine ion (CIx = 3.00 x 10-9 meters. Assume that this x value is muchlarger than the separation d between the charges in the dipole,so that the approximate expression for the electric field along thedipole axis can be used.Part Aof charge -1.60 x 10-19 C, is located atFind the magnitude of the electric force, ignoring the sign, that the water molecule exerts on the chlorine ion.• View Available Hint(s)6.57 • 1012N

Dipole moment of Water, pw = 6.17×10-30 C.mCharge on Chlorine ion, qCl = -1.6×10-19 CDistance, x =…

Answered: The dipole moment of the water molecule…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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The dipole moment of the water molecule (H,O) is
6.17 x 10-30 C.m. Consider a water molecule located at the
origin whose dipole moment p points in the positive x direction. A
chlorine ion (CI
x = 3.00 x 10-9 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.
Part A
of charge -1.60 x 10-19 C, is located at
Find the magnitude of the electric force, ignoring the sign, that the water molecule exerts on the chlorine ion.
• View Available Hint(s)
6.57 • 1012
N

Expert Solution

Step 1

$D i p o l e m o m e n t o f W a t e r, p_{w} = 6.17 \times 10^{- 30} C . m C h a r g e o n C h l o r i n e i o n, q_{C l} = - 1.6 \times 10^{- 19} C D i s \tan c e, x = 3 \times 10^{- 9} m C o u l o m b' s c o n s \tan t, k = 8.99 \times 10^{9} N m^{2} / C^{2}$

We assume that x is much larger than the distance d between the charges in dipole.

So, for electric field we have,

$E = \frac{2 k p}{x^{3}}$

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