# Gauss's Law (i think)

## Homework Statement

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.

## Homework Equations

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?

Welcome to the forum.

Why do you take Q= pV ?
What did you take for r? In the Electric field?

Q = enclosed charge?
p = charge density & V = volume, so the units are just C when you multiply pV

for r, i set whatever the volume is and i set it to equal (4/3)pi(R^3) and solved for R

nevermind i figured how to do it!

i used..
E = kQ/(r^2)
Q = (# protons)(elementary charge)