Determine the magnitude of the electric field at the surface of a lead-199 nucleus, which contains 82 protons and 117 neutrons. Assume the lead nucleus has a volume 199 times that of one proton and consider a proton to be a sphere of radius 1.20 10-15 m.
E = kQ/r^2
Q = pV
p = Ne = (#protons)(1.6x10^-19)
V = Volume Nucleus = (#lead)(4/3)(pi)(r^2)
The Attempt at a Solution
tried plugging all the numbers in and came out with a wrong answer..
is my method correct?
is there an easier way such as enclosing the atom in a Gaussian sphere?
..making E = sigma/epsilon_naught
what would sigma be in this case?