Spherically symmetric charge density given electric potential

In summary, a spherically symmetric charge distribution can be described by an electric potential of the form E = - ∇ V, where the gradient is in spherical coordinates. By calculating the electric field using the first equation and plugging it into the difference equation with r = r and r' = (r+dr), one can solve for p(r) to determine the charge density. However, this may result in a large number of variables and it is recommended to neglect higher-order terms and only consider terms proportional to dr. Additionally, setting r'=0 and integrating the right side of the equation from 0 to r can simplify the solution and result in Gauss' Law, allowing for an algebraic solution for \rho.
  • #1
stauber28
12
0
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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  • #2
Did you at least calculate the electric field?

E = - ∇ V where the gradient is in spherical coordinates.
 
  • #3
I calculated E using the first equation. E=dV/dr
 
  • #4
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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  • #5
You have the right approach. You can neglect the higher-order terms. Keep only the terms proportional to dr.
 
  • #6
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 [itex]\rho[/itex].
 

FAQ: Spherically symmetric charge density given electric potential

What is spherically symmetric charge density?

Spherically symmetric charge density refers to a type of charge distribution in which the charge is evenly distributed around a central point, resulting in a symmetrical shape.

How is spherically symmetric charge density related to electric potential?

The electric potential at any point in a spherically symmetric charge distribution is determined by the amount of charge and its distance from the center. This relationship is described by Coulomb's law.

Can spherically symmetric charge density be represented mathematically?

Yes, spherically symmetric charge density can be represented mathematically using the charge density formula, which is equal to the charge per unit volume.

What is the formula for calculating electric potential from spherically symmetric charge density?

The formula for calculating electric potential from spherically symmetric charge density is V = kQ/r, where k is the Coulomb constant, Q is the total charge, and r is the distance from the center.

What are some real-life examples of spherically symmetric charge density?

A few examples of spherically symmetric charge density in real life include a charged sphere, a metal ball with a net charge, and an electrically charged planet or star.

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