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nrqed
#6
Mar29-08, 07:03 PM
Sci Advisor
HW Helper
P: 2,953
Quote Quote by ~christina~ View Post
yes, I did mean that, but since I copied and pasted here and there I missed that.

[tex]F= 6.67x10^{-11}N(m/kg)^2 \frac{(6.60 x10^ {-27}kg)
(3.25x10^{-25}kg)} {(9.1x10^{-15})^2= 1.727x10^-33 N}[/tex]




hm so you mean I take this number below and subtract it from the distance if it was 2d? like this?

[tex]U= (8.9876x10^9 N*m^2/C^2) \frac{|3.2x10^{-19} C ||1.3 x10^{-17} C |} {(9.1 x10^{-15}m)} = 4.1086x10^{-12} [/tex]

[tex]U= (8.9876x10^9 N*m^2/C^2) \frac{|3.2x10^{-19} C ||1.3 x10^{-17} C |} {(2(9.1 x10^{-15}m))} =2.054x10^{-12} [/tex]
Yes (but don't put absolute values...there are no absolute values taken when we calculate potential energy. here it does not make any difference but just to let you know).

And notice that the units are Joules. You will have to convert to MeV (recalling that [tex] 1 eV = 1.60 \times 10^{-19} J [/tex])