# Inputting complicated equation into Wolfram

I am having a bear of a time getting this equation into Wolfram so I can solve it for E(r) = 1000.

Everything is constant except for little r, which is the cursive r in each term. In case anyone is wondering, this is the expression for the electric field from the center of a hydrogen atom. r is for radius.

Can anyone put this in and link me to the solution? Much appreciated.

• I cannot read all of it. Is that an $M$ or $\pi$? – Amzoti Oct 31 '13 at 2:50
• @Amzoti I think (after staring for a bit) all of the things that look like m/$\pi$ are actually $r$, and that is what OP wants to find. After more staring, I would not swear by it. – J. W. Perry Oct 31 '13 at 2:53
• @J.W.Perry: Thanks, for a sec I thought, man I am getting old! :-) – Amzoti Oct 31 '13 at 2:55
• The m-like things are actually lowercase r's. The only pi is between the 4 and epsilon naught. – user99984 Oct 31 '13 at 2:57

Here is one approach, but you can put in your constants and then play around with defining them using the proper names.

  solve (.2)/(4 pi .5 r^2)(1 - e^(-2 r /.6)(2(r/.6)^2+2(r/.6)+1)) = 1000 for r


See this WA page.

By defining parameters, I mean, something like:

 q = .2, solve (q)/(4 pi .5 r^2)(1 - e^(-2 r /.6)(2(r/.6)^2+2(r/.6)+1)) = 1000 for r


Here you can define your parameters:

q=.2, t = .4, a=.6, solve (q)/(4 pi t r^2)(1 - e^(-2 r /a)(2(r/a)^2+2(r/a)+1)) = 1000


For the parameters you posted, I got two solutions:

• $\large r = 7.71723 \times 10^{-20}$
• $\large r = 1.19945 \times 10^{-6}$

Note, there are other options:

• Get a free CAS like SAGE, Maxima or others. These also have working online copies.
• Use Mathics.org
• Can you solve this for me with q = 1.6*10^-19, epsilon nought = 8.85*10^-12, a = 0.529 * 10^-10. I'm having trouble defining the constants and still getting it to solve. – user99984 Oct 31 '13 at 3:15
• I got an answer, which was r = 4.76 * 10^-15 m – user99984 Oct 31 '13 at 4:25
• It is possible I am doing something wrong with the solver, but currently do not see what that is. What did you use? – Amzoti Oct 31 '13 at 4:27
• My TI 83. I think I can explain what's going wrong with your solver. The function is actually increasing for extremely small values of r, then decreasing for all other ones. Based on some other calculations I did, it should be around the order of 10^-17. – user99984 Oct 31 '13 at 4:59
• Yes, yesterday was fun - seeing all the little "goblins", and a few not so little goblins! – Namaste Nov 1 '13 at 12:06