Calculating Electric Field at Lead-199 Nucleus with Gauss's Law

[physizl]

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?

 

[betel]

Welcome to the forum.

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

 

[physizl]

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

 

[physizl]

nevermind i figured how to do it!

i used..
E = kQ/(r^2)
Q = (# protons)(elementary charge)
r = [(# lead)^(1/3)](radius proton)

plugged Q and r back in E and came out with the correct answer

 

[paranormal]

I would like to provide a more detailed response to this problem. Firstly, Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed within that surface. It can be used to calculate the electric field at a point outside a charged object by considering the charge enclosed within a Gaussian surface surrounding that point.

In this problem, we are trying to calculate the electric field at the surface of a lead-199 nucleus. To do this, we can enclose the nucleus in a Gaussian sphere with a radius equal to the radius of the nucleus (since we are only interested in the electric field at the surface). The charge enclosed within this sphere is the sum of the charges of all the protons in the nucleus, which can be calculated using the formula Q = pV, where p is the charge density (in this case, the charge of a proton) and V is the volume of the nucleus.

So, the charge enclosed within the Gaussian sphere is Q = (#protons)(1.6x10^-19)(4/3)(pi)(1.20x10^-15)^3. Plugging in the values of 82 protons and the given radius, we get Q = 2.03x10^-16 C.

Now, using Gauss's Law, we can calculate the electric field at the surface of the nucleus by dividing the charge enclosed by the surface area of the Gaussian sphere, and then multiplying by the appropriate constant. So, E = (kQ)/(4(pi)(r^2)), where k is the Coulomb constant and r is the radius of the Gaussian sphere. Plugging in the values, we get E = (8.99x10^9)(2.03x10^-16)/(4(pi)(1.20x10^-15)^2) = 4.76x10^12 N/C.

Therefore, the magnitude of the electric field at the surface of the lead-199 nucleus is 4.76x10^12 N/C.

In terms of an easier method, using the formula E = sigma/epsilon_naught, where sigma is the surface charge density and epsilon_naught is the permittivity of free space, can also be used to calculate the electric field at the surface. However, in this case, we do not know the surface charge density of the nucleus. We can only calculate the total