The dipole moment of the water molecule (H2O) is 6.17x10^-30C.m. Consider a wate
ID: 1273481 • Letter: T
Question
The dipole moment of the water molecule (H2O) is 6.17x10^-30C.m. Consider a water molecule located at the origin whose dipole moment p points in the +x-direction. A chlorine ion (Cl-), of charge-1.60x10^-19C, is located at x=3.00x10^-9m. Assume that x 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
Find the magnitude of the electric force that the water molecule exerts on the chlorine ion.
Part B
What is the direction of the electric force.
Part C
Is this force attractive or repulsive?
Explanation / Answer
The dipole moment = p = 6.17x10^-30 C.m.
The dipole is located at the origin
Dipole moment p points in the +x-direction
Charge on chlorine ion = q = -1.60x10^-19C ,
Chlorine ion is located at x=3.00x10^-9m .
Assuming 'x' is much larger than the separation d between the charges in the dipole,
The approximate expression for the electric field along the dipole axis =E = 2kp/x^3
The magnitude of the electric force exerted by the water molecule on the chlorine ion=F =qE
The magnitude of the electric force exerted by the water molecule on the chlorine ion = F =q[2kp/x^3]
The magnitude of the electric force exerted by the water molecule on the chlorine ion = F = (1.60x10^-19)*2*9*10^9[ 6.17x10^-30 ] /( 3.00x10^-9 )^3
A)The magnitude of the electric force exerted by the water molecule on the chlorine ion = F = 6.5813*10^-13 N
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As chlorine ion is negativr force is attractive , in negative x- axis direction
B)The direction of the electric force is -x-direction
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C) This force is attractive
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