Correct, computer gets: BDFG Hint: The electric field is a vector quantity (with

Question

Correct, computer gets: BDFG Hint: The electric field is a vector quantity (with direction and magnitude), while the electric potential is a scalar quantity. The formula for electric potential (also referred to as the "voltage") is given by Equation 17-5 in your textbook. You won't need to do much (any?) computing here, you just need to think about directions, signs, and what terms might be bigger than others... (Note that when we have collections of point charges like this, we usually make the implicit assumption that the zero of Voltage is off at infinity, that's built in to Eq 17-5 of the text) 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 = -1.024 times 10-19 C, the angle at the oxygen atom is = 103.9 degree and the distance between the oxygen and the hydrogen atoms is d = 97.4 times 10-9 m. What is the potential at the midpoint between the two hydrogen atoms? An electric field of 680 V/m is desired between two parallel plates 17.8 mm apart. How large a voltage should be applied? Correct, computer gets: 1.21e+01 V Hint: See section 17.2 of your text. (This question is really just asking for the magnitude of the voltage difference, your answer should be positive...)

Explanation / Answer

Potential is a scalar quantity so you wouldn't add it up like electric force. Instead find the potential at that spot between each charge (the sign of the charge determines the sign on the potential) and add them together. Potential = V= kQ/r.

First we'll find the potential from Q1: the distance r is d*cos(?/2)= 97.4e-9m* cos(103.9/2) = 6.0e-8m.

Then plug that into V=kQ1/r = 8.99e9*-1.024e-19C/6.0e-8m= -0.015V

Second find the potential of both Q2 (they are the same) with distance r as

d*sin(?/2)= 97.4e-9m*sin(103.9/2)= 7.7e-8m

You know the charge of Q2 is positive and half Q1 since the molecule is neutral:

Q2 = 1.024e-19C/2= +5.12e-20C

Then plug that into V=kQ2/r = 8.99e9*5.12e-20C/7.7e-8m = +0.00598V

Add the potentials together to get your net potential

V= VQ1 + VQ2 +VQ2 =-0.015V + .00598V + .00598V = -.003V

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