Dielectric problem

Can you give us a bit of background on this question, Emily, perhaps start by telling us what textbook you got it from? This will help us answer it. It's been a while since I've done chemistry though I'm willing to give it a shot.

It's from a worksheet. Here's the whole problem:

Give the current state of the economy, lots of Maxwell's Demons are unemployed and "will push molecules for food." Accordingly, I hire a couple of demons to hold a Calcium ion (Ca^+2) 6 Angstrom (0.6nm) from a sulfide ion (S^2-)
How much energy do they have to exert to keep the ions in this position? Assume the demons are working in a vacuum.

I know you use this eqn: Use U=k (q1q2)/ Er but I got stuck halfway

I'll attempt to answer this now I that I've figured out that we're using coulombs law.

U = k(q1)(q2)/ r2

http://upload.wikimedia.org/math/e/1/6/e16f4b0bf23cd9354b492e3581cf9d0f.png
 
Where U = potential energy
k = constant for a medium (medium is one in which charges are kept, i.e. air, water, etc.)
k also equals to 1/4(pi)(eo) which should be given.
q1, q2 = charges (including their sign, positive or negative)
r = distance between charges

U = (8.99 x 10^9 nm)(2)(-2) / 0.6 nm2

1 Angstrom = 1.0 × 10-10 meters

By substituting in your k value, you can obtain U.

Tell me if you get the right answer after plugging in all those numbers.

Would it be U= (k)[ (q1q2) / Er] I think you forgot the E (dielectric constant) and I believe 8.99x10^9 is supposed to be in JmC2-
I dont have the answer, but I tried your method for this other problem which I have the answer for and it didn't match with what they got.

Inside a protein, a ferric (Fe+3) ion is separated from a cupric ion (cu+2) by a distance of 1.5 nm. What is the energy of attraction between the two ions? Assume the dielectric constant in the protein environment is the same as for water which is 78.54.
Answer: 6948J/mol

I'm not sure It should be as straight forward as plugging in the values I found this question and it's pretty much the same, and all they did was plug in the values given in the question.

Calculate the force of attraction between a Ca2+ ion and an O2- ion, the centres of which are 1.25 nm apart.

Answer

This just involves plugging numbers into the formula

F = q1q2/(4 pi epsilon r2)

The numbers to plug in are: q1 = q2 - 3.2 * 10-19 coulombs, r=1.25 * 10-9 m and epsilon = 8.85 * 10-12 F/m, giving the result F = 5.89 * 10-10 N.

Sorry Emily. I'm not much of a physics person so hopefully this example gives you some insight. If you do end up solving, I'd like to know how for future reference.

Is this problem solved?