Prerequisites:

• Water-coat forces
  
• Energy of place -- potential energy
• Electric potential energy
• Integrals
• Newton 2 on a spreadsheet

In the problem, Water-coat forces, we analyzed the force between an isolated ion (e.g., Na+ or K+), and a water dipole. Although the water molecule is neutral, since all its charges are not in the same place, when oriented properly, it is attracted to the ion. We found the formula for the force on a water-like dipole as a function of the distance between the ion and the dipole. The result we found was the following formula:

The distances a, b, and x are as shown in the figure. The parameters are as follows:

• k = 9 x 10⁹ N-m²/C² (Coulomb's constant)
• e = 1.6 x 10^-19 C (charge of an electron)
• a = 60 nm (height of water molecule triangle)
• b = 75 nm (base of water molecule triangle)

In this problem, you will explore how this force creates a potential energy between the ion and the water molecule.  And that even though all the forces we are adding go like 1/r², the result does not. You will show this using a numerical calculation on a spreadsheet and see whether a power law is a good approximation for the potential energy as a function of distance (and for the force).

Download the spreadsheet Ion-DipolePotential.xlsx. The structure of the sheet has been set up for you including places for the parameters (and the parameters are named so you can use them in formulas on the spreadsheet -- see the discussion in Solving Newton 2 on a Spreadsheet if you forget how this works). And the graphs are even set up.  All you have to do is write the equations. This is not so trivial so let's step through what to do.

Since our goal is to construct a potential energy, we want to add up the work done by the force as the objects get closer together. So we'll start at some value far out (x0) and step inward by small steps (dx). We have to be careful not to get too close because the formula has a 1/x².  If the x gets too small, the potential gets huge! (Really, this never happens because when atoms get close enough to start to overlap, quantum effects produce a repulsion.) The parameters of the calculation are set out in the spreadsheet as shown below:

The values of the parameters are put into cells B4 through B10. (Cell B8 is calculated -- the rest are entered as numbers.) We have chosen the following units in order to make the numbers of a reasonable size: force in attoNewtons (10-18 N) charge in nanoCoulombs (10-9 C) distance in nanometers (10-9 m).   The cells have been given the following names so you can use these names in formulas (without worrying about the row and column numbers -- and then when you automatically extend or copy a formula, the cells those refer to don't change. The cell names that you can use in you formulas are as follows: Coulomb coefficient = 2ke2: CCoeff Water size parameters a and b: a, b Starting distance for integrating inward: x0 Step size for integrating inward: dx

A. Plotting the force

To get a plot of the force as a function of distance, note that the column of positions, x, is already filled out for you. The first value is x0, and each succeeding value represents the equation,

x_new = x_old - dx

(We have a minus sign since we are working our way in to smaller values of x.) Make sure that you understand how the formulas in these cells work (by cells we mean the cells in the excel software, not biological cells).  Now enter the spreadsheet translation of the messy formula  of the force from above into cell B13, using the cell label, A13, to replace x.  You might want to put in a lot of parentheses to make sure that this messy formula is correctly calculated by the spreadsheet. Now copy the formula into other other cells below by putting the cursor on the lower right corner of the box (so it turns from an open cross to a solid cross) and double click to fill the formula down the column -- or just drag that cross down to the bottom of the column.

You should now have a plot of F in one of the graphs to the right.

B. Fitting the force

Our formula is a mess.  It has a term that looks like 1/distance² as we expect, but there are two terms like that and they tend to cancel at long distances. How well is it represented by 1/x²? Or by 1/x to any other power N  (in other words does the force change as distance in the following way:  1/x^N ) ? To see this, we have set up parameters and columns to create a force that looks like 1/x to some power.  The formula we will use to try to fit the force with a power law is

Power Law Force = A/x^N

where A and N are constants that we will adjust. The value of A for the force is in cell F5 and has the name AF.  The value of N for the force is in cell F6 and has the name NF. 

The constant A have been calculated so that it matches the real value of F at the starting point; that is (in what is below we express both the force F and the power law force PLF as functions of position)

PLF(xstart) = F(xstart)

so

AF/(xstart)^NF = F(xstart)

or

AF = F(xstart) * (xstart)^NF.

You can see that the fit is not too good.  It seems like a simple power law is NOT describing our messy formula. 

Adjust the power of N in the PLF law (cell F6) until you get the best fit you can for the range from 150 to 100 nm.  What happens if you look closer in -- from 100-to 50?  What are the best fit powers for these cases? Print the graphs and include them in your homework.

Another way to discover the functional dependence of a formula is to use log-log plots, where both the horizontal and vertical axis are logarithms of the variables. These allow you to distinguish different "power laws".  

Try this:  Calculate the logarithm of the position, and the logarithm of the force in the columns provided.  You should see a plot of log(F) vs log(x).  Is the plot of log(F) vs log(x) linear?  What is the slope (you can fit a trendline in Excel) What does the slope of this "log-log plot" mean? 

C. Calculating the potential

We'll now calculate the potential using the formula

ΔU = -Fdx.

The change in the PE is the negative of the work done by the force. Since the force changes with x, we have to do this in small steps. We can do this numerically by opening up the delta:

U_new = U_old -F dx

To get a starting value we should really go very far out, start at 0, and work our way in. If we did this, we would get a non-trivial value of U at x = 150 nm.  Some exploration suggests that a good starting value for U at that distance is about -880. Put this value in the first box in the column marked U , and then in the next box down (the new U), put the cells to make the above formula. Now copy this down.  You should get a graph of the potential in the other graph at the right. Does it behave like you expect? Is it positive or negative? Explain how you know what it should do. If it doesn't do that, fix it!

D. Fitting the potential

Now we'll do the same thing with U that we did with F -- we'll see if the curve generated by the integration (Yes -- that's what you were doing by adding up all those little pieces. An integral!) is fit by 1/x or by something else. Again, now an attempt to fit U with a Power Law Potential (PLU) has us compare our result to a formula

PLU = AU/x^NU

We'll fit the coefficient AU so that the power law potential fits at the first point, and then adjust the parameter NU (in cell F9) to get the best fit you can.

Does the power you get change as you move in -- to 100?  (Careful! You have to take the starting value for U from the calculation from 150 to 100.)