Question
http://jilawww.colorado.edu/beckergroup/figures/Water.jpg
Let's consider once again the water molecule with two hydrogen atoms and one oxygen atom. Since the oxygen atom has a higher electronegativity than the hydrogen atom, the side of the molecule with the oxygen atom has a partial negative charge. Assume that the partial negative charge on the oxygen atom is Q1 = -1.056 x 10-19 C, the angle at the oxygen atom is = 104.2o and the distance between the oxygen and the hydrogen atoms is d = 92.2 x 10-9 m. What is the potential at the midpoint between the two hydrogen atoms? Hint:There are potentials from three single point charges, which have to be added. Draw a careful picture, use the geometry to figure out the relevant distances. (Remember that voltages can be positive or negative!)
Explanation / Answer
Q2=1.6*10^-19C (charge of proton)
V=KQ1/dcos(theta/2) + 2*kQ2/dsin(theta/2)
=-9*10^9*1.056 * 10^-19/(92.2* 10^-9*cos52.1) + 2*9*10^9*1.6 * 10^-19/(92.2* 10^-9*sin52.1)
=0.0228 V.
