Mastering Physics Solutions: Electrostatic Force of Water on a Chlorine Ion

Electrostatic Force of Water on a Chlorine Ion

Part A = 6.58 * 10^-13 N
Part B = negative x
Part C = attractive

Solution Below:

The dipole moment of the water molecule (H2O) is 6.17 * 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 (Cl-, of charge -1.60 * 10^-19 C, is located at x = 3.00 * 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

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

Start by finding the electric field, E. The electric field can be approximated by E = 2kp/x^3:

E = 2kp/x^3
E = 2*9*10^9[ 6.17x10^-30 ] /( 3.00×10^-9 )^3
E = 4.11 * 10^6

Now find the magnitude of the force, which is just F = qE

F = qE
F = (-1.60 * 10^-19) * (4.11 * 10^6)
F = 6.58 * 10^-13 N

6.58 * 10^-13 N

Part B

What is the direction of the electric force?

  • negative x
  • positive x

negative x

Part C

Is this force attractive or repulsive?

  • attractive
  • repulsive


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