8.2.P14
To try to get some understanding of how a water molecule looks to an ion electrically, let’s map out a piece of the electric potential produced by the charges in water. We’ll use a toy (easy to calculate model) of water: Two positive charges of +e (hydrogen) and a negative charge of 2e (oxygen) with the bonds at a right angle as shown in the figure. Take the separation of each + from the – charges as the bond length d. (Just to get a sense of scale, real water molecules have a value of d ~ 0.1 nm. We won't use this in this problem. Instead, we will construct dimensionless forms of the potential to study.)
A. Calculate the electric potential, V(A), that a test charge would feel if placed at the point A. Express your answer as a number times the combination of symbols, k_{c}e/d.


The contribution of each charge q to the potential at any point of interest is kCq/r where r is the distance from the charge to the point of interest. If we take A to be our point of interest, we have two charges, +e and +e a distance r = d/√2 away, and a charge 2e, also a distance r = d/√2 away (using the Pythagorean theorem). Our result is therefore
V(A) = k_{C}e/r + k_{C}e/r  2k_{C}e/r) = k_{C}e/r [1 + 1  2] = 0 (k_{C}e/d)
B. Calculate the electric potential, V(B), that a test charge would feel if placed at the point B. Express your answer as a number times the combination of symbols, k_{c}e/d.
The positive charges are a distance d from B and by the Pythagorean theorem B is a distance d√2 from the double negative charge. Our result is therefore
V(B) = k_{C}e/d + k_{C}e/d  2k_{C}e/(d√2)) = k_{C}e/d [1 + 1  2/√2] = 0.59 (k_{C}e/d)
C. Write an equation to express the electric potential, V(x) that a test charge would feel if placed at a position x on the positive xaxis. By dimensional analysis, we know that the final equation must have the form
V(x) = (k_{C}e/d) f(X)
where X is the dimensionless function of the dimensionless variable, X = x/d. Write your result in this form by using legitimate algebraic manipulations to identify the function f(X).
To get the equation it would help to have a labeled picture. The distance of the point of interest (x,0) (black dot) from the positive charges is r^{2} = (d/√2)^{2} + (xd/√2)^{2} or
r^{2} = d^{2}/2 + x^{2} + d^{2}/2 2xd/√2
Simplifying, this becomes
r^{2} = x^{2} + d^{2}  xd√2


Since we are going to want to put this into dimensionless form, let's change to our dimensionless variable X by replacing x with x = Xd. This gives
r^{2} = X^{2}d^{2} + d^{2}  Xd^{2}√2 = d^{2}(X^{2} + 1  X√2)
This is a bit of a mess. Let's simplify our manipulations by defining the dimensionless function
g^{2}(X) = X^{2} + 1  X√2.
Then we get r = d g(X).
So our potential is
V(x) = 2k_{C}e/r  2k_{C}e/x
Since x = Xd, the second term is 2kCe/(dX). So the result is
V(x) = k_{C}e/d {2/g(X)  2/X}
so the function that gives us the shape of the potential is
f(X) = 2/g(X)  2/X
with
g(X) = {X^{2} + 1  X√2}^{1/2}
D. Use a spreadsheet or graphing calculator to make a plot of the function f(X) (for positive X) and include this plot in your assignment. Identify qualitative features of your function (Does it go to 0 anywhere? Does it go to infinity anywhere? Does it have a peak anywhere?) and explain whether these are reasonable given the physical situation.
Using the Desmos graphing calculator, we get the following graph:
Even if we didn't have the explicit function to plot, this shape is what we should expect. As x gets small, the test charge approaches the 2e charge and the strong attraction (and infinite negative potential at x=0) should dominate. At large positive distances, since the positive charges are a bit closer, we expect that we should get a positive potential. Since the net charge of the water molecule is 0, we expect that the potential should fall off like a dipole  like 1/x^{2} rather than like 1/x  but should go to 0 as the distances get large.
E. If a positive ion approached the molecule along the x axis from the right, would it be attracted or repelled from the molecule? Is this consistent with the information in your graph? Explain.
If we are right between the two positive charges, the force from them will cancel so we will be left with an attraction to the negative charge. As we go farther out, we expect the positive charges to be more important. This shows up in our potential curve in that at distances greater than x = d (this is X=1) the curve is decreasing. It has negative slope so the E field (which is = dV/dx) is positive and would repel a positive test charge.
Joe Redish 2/20/20 solution 2/21/20
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