The O-H bond length in a water molecule is 0.958 A (1 A = 1010 m). The oxygen en

Question

The O-H bond length in a water molecule is 0.958 A (1 A = 1010 m). The oxygen end of the molecule is a little more negative than the hydrogen end. The result is that the oxygen has an effective charge of 0.644e and the hydrogen has an effective charge of +0.644e. What is the effective electric force between the oxygen and hydrogen atom? Calculate the magnitude and describe the direction of the force. The Coulomb’s Law constant is k = 8.99 × 109 N·m2/C2 .

The answer is: F = 1.04 X 10^-8 N with direction attractive between the oxygen and hydrogen, what I would like to know is how to get the answer. I'm using the equation F=KQ(1)Q(2)/d^2 but I am not getting the right answer. Please help!

Explanation / Answer

Given,

r = 0.958 x 10-10 m ; q1 = -0.644 6 = 0.644 x 1.6 x 10-19 C = - 1.03 x 10-19 C

q2 = + 1.03 x 10-19 C

We need to find the electrostatic force between oxygen and hydrogen atom.

We know from Coulomb's law that the electric force between two charged particle is given by:

F = k q1 q2 / r2

F = 9 x 109 x 1.03 x 10-19 C x 1.03 x 10-19 C / (0.958 x 10-10 m)2 = 1.04 x 10-8 N

Hence, F = 1.04 x 10-8 N

Since the chares are opposite, the force between them will be attractive.

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