Kas0988
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Homework Statement
What is the magnitude of the force on an electron at a distance of 1.79 Angstrom from the radium nucleus?
Homework Equations
Coulomb's Law:
<br /> F = \frac{1}{4 \pi \epsilon_0} \ \frac{q_1 q_2}{r^2}<br />
The Attempt at a Solution
For this problem, I first rearranged Coulomb's Law to the form of (k*q1*q2)/r2. The variables where then as followed:
k = 9e9 Nm2/C2
q1 = -1.6e-19 C
q2 = 1.6e-19 C
r = 1.79e-10 m
k is being used as our constant, while q1 and q2 are the charge of an electron and proton respectively. r is our distance, which has been converted from Angstroms to Meters. By plugging all of this in, it yields the following:
[(k = 9e9 Nm2/C2)*(q1 = -1.6e-19 C )*(q2 = 1.6e-19 C)]/(1.79e-10m)^2
[-2.30400000000e-28]/[3.20410000000e-20]
Answer: -7.19101123596e-09 N
This answer is not correct for this problem, however. I suspect that either my calculations yielded an error, I forgot some type of conversion, I am missing a step, or I did not take into account some special step that involves the nucleus of Radium. I'm flatout stuck at this point though, and don't have a clue what to do from here on out.