The dipole moment of the water molecule H2O is 6.17x10^-30rm C*m. Consider a wat
ID: 1691743 • Letter: T
Question
The dipole moment of the water molecule H2O is 6.17x10^-30rm C*m. Consider a water molecule located at the origin whose dipole moment vector p points in the positive x direction. A chlorine ion 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. A. Find the magnitude of the electric force, ignoring the sign, that the water molecule exerts on the chlorine ion. B. What is the direction of the electric force? (positive or negative x direction)C. Is this force attractive or repulsive? The dipole moment of the water molecule H2O is 6.17x10^-30rm C*m. Consider a water molecule located at the origin whose dipole moment vector p points in the positive x direction. A chlorine ion 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. A. Find the magnitude of the electric force, ignoring the sign, that the water molecule exerts on the chlorine ion. B. What is the direction of the electric force? (positive or negative x direction)
C. Is this force attractive or repulsive?
Explanation / Answer
The dipole moment of water was given, not its charge. Dipole moment p=qd so you have to first plug in and solve to find water's charge. You get 6.17 X 10^-30 = q (3 x 10^-9), solving for q gives water's charge 2.0566 x 10^-21. Now you can plug both charges into the equation E=k(qq)/r^2 and you get 3.28 x 10^-13. Mulitply that by 2 and you get 6.58 x 10^-13 which is the correct answer.