Spherically symmetric charge density given electric potential

AI Thread Summary
A spherically symmetric charge distribution leads to a specific form of electric potential, prompting the need to determine the charge distribution. The electric field can be calculated using E = -∇V in spherical coordinates, with a focus on the difference in electric field between two radial points. The discussion highlights the challenge of simplifying the resulting equations to express charge density, with suggestions to neglect higher-order terms and integrate from 0 to r. By applying Gauss' Law, one can algebraically solve for the charge density, ρ. The conversation emphasizes the importance of careful manipulation of variables and integration techniques in deriving the desired form.
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1. A spherically symmetric charge distribution results in an electric potential of the form
pt1.jpg


What is the charge distribution?

2.
Hint: consider the difference in electric field between two values of r
pt2.jpg


Show that the answer is of the form
pt3.jpg



3. I have attempted several solutions but haven't gotten anywhere.
 
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Did you at least calculate the electric field?

E = - ∇ V where the gradient is in spherical coordinates.
 
I calculated E using the first equation. E=dV/dr
 
Correction E=-dv/dr <-- I am unsure to just have dr=dr or dr = (the derivative of everything in the 1st equations brackets[] ) - I decided to just have it equal dr

I then plugged the E value into the difference equation with r = r, and r' = (r+dr). I took the difference and set it equal to the right side of the same equation, and then solved for p(r). I hoped this would give me something in the form of charge density noted. However, I ended up with a large number of variables which would not simplify to this form.

Any Insights?
 
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You have the right approach. You can neglect the higher-order terms. Keep only the terms proportional to dr.
 
You can have r'=0 and just integrate the right side of the equation from 0 to r. You now have Gauss' Law, and you can just solve algebraically for \rho.
 
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