The dipole moment of the water molecule is 6.17 * 10^-30 . Consider a water mole
ID: 1706142 • Letter: T
Question
The dipole moment of the water molecule is 6.17 * 10^-30 . Consider a water molecule located at the origin whose dipole moment points in the +x-direction. A chlorine ion , of charge , -1.6 * 10^-19 is located at: x= 3.00 * 10^-9m. Assume that x is much larger d than the separation between the charges in the dipole, so that the approximate expression for the electric field along the dipole axis can be used.
I found that the force was 6.58*10^-13 N.
B) What is the direction of the electric force?
a. -x-direction or b. +x-direction
C) IS this force:
a. attractive or b. repulsive
Explanation / Answer
Given that The dipole moment of the water molecule si , p = 6.17 *10^-30 The charge of chlorine ion, q = -1.6 *10^-19C The calculated force is 6.58 *10^-13 N Now, we will find the electric field along the dipole is E = F/q = (6.58 *10^-13N)/1.6 *10^-19 C E = 4.11 *10^6 N/C Since, the dipole moment is positive, the electric field is also positive. B) Here the charge is negative ,so the direction of electric field and the force will be opposite.Hence, the direction of the electric force is towards -x direction. option b is correct C) Since the force is negative, it is an attractive force option a is correct