• Support PF! Buy your school textbooks, materials and every day products Here!

Charge density for a given potential

  • Thread starter Silviu
  • Start date
  • #1
624
11

Homework Statement


Given the electric potential ##V(r)=A\frac{e^{-\lambda r}}{r}## calculate the charge density ##\rho(r)## and the electric field ##E(r)##.
They specify the answer for charge density should be: ##\rho = \epsilon_0 A(4\pi \delta^3(r)-\lambda^2e^{-\lambda r}/r)##

Homework Equations


##E=-\nabla V##
##\nabla E = \frac{\rho}{\epsilon_o}##

The Attempt at a Solution


For the electric field I got: ##E = Ae^{-\lambda r}(1+\lambda r)/r^2##, pointing in the r direction and this is the same as their answer. Then I tried to take the divergence of E, and in spherical coordinates this would be ##\nabla E = \frac{1}{r^2}\frac{\partial r^2 E}{\partial r} = \frac{1}{r^2}\frac{\partial (Ae^{-\lambda r}(1+\lambda r))}{\partial r} ## . However this doesn't give me a delta function so I am not sure where I did something wrong.
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,547
1,155
Your expression for the divergence is only valid for ##r\ne 0##. You need to consider what's happening at ##r=0## in another way.
 
  • Like
Reactions: BvU
  • #3
624
11
Your expression for the divergence is only valid for ##r\ne 0##. You need to consider what's happening at ##r=0## in another way.
Thank you! So I found a solution online to this problem but I am still a bit confused. I attached their solution. So i understand how they obtain the delta function but when they apply the divergence to the other term, they use ##\frac{\partial}{\partial r}##. Shouldn't they use the divergence in spherical coordinates which is ##\frac{1}{r^2}\frac{\partial (r^2 E')}{\partial r}##, with E' being the term that remained to be acted on?
 

Attachments

  • #4
TSny
Homework Helper
Gold Member
12,300
2,794

Related Threads for: Charge density for a given potential

Replies
2
Views
953
Replies
6
Views
890
Top