Thread: Particles and charge View Single Post
 Quote by ~christina~ yes, I did mean that, but since I copied and pasted here and there I missed that. $$F= 6.67x10^{-11}N(m/kg)^2 \frac{(6.60 x10^ {-27}kg) (3.25x10^{-25}kg)} {(9.1x10^{-15})^2= 1.727x10^-33 N}$$ hm so you mean I take this number below and subtract it from the distance if it was 2d? like this? $$U= (8.9876x10^9 N*m^2/C^2) \frac{|3.2x10^{-19} C ||1.3 x10^{-17} C |} {(9.1 x10^{-15}m)} = 4.1086x10^{-12}$$ $$U= (8.9876x10^9 N*m^2/C^2) \frac{|3.2x10^{-19} C ||1.3 x10^{-17} C |} {(2(9.1 x10^{-15}m))} =2.054x10^{-12}$$
And notice that the units are Joules. You will have to convert to MeV (recalling that $$1 eV = 1.60 \times 10^{-19} J$$)