Finding the potential of a charged, solid sphere using the charge density

In summary, the conversation discusses finding the potential at the center of a solid ball of charge with a given radius and volume charge density. The individual has already calculated the electric field and potential using a formula, but is now trying to find the potential using the charge density. They are unsure of the first step and whether their proposed method is correct. They receive confirmation that their method is valid and discuss the specifics of the integral and volume element. Ultimately, they arrive at a potential value that they are satisfied with.
  • #1
LionelHutz
5
0

Homework Statement



Solid ball of charge with radius R and volume charge density ρ(r) = ρ0r2, centred at the origin.

I have already found the electric field for r<R and r>R and also the potential at the origin by using the formula:
V = -∫E.dl

Now i want to find the potential at the origin using the charge density but am at a loss for the first step

Homework Equations



V = (1/4∏ε)∫(ρ/r)d[itex]\tau[/itex]


The Attempt at a Solution



I don't need the full solution, just the first step would be much appreciated.
 
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  • #2
Find the potential at the center due to a thin shell of charge and then integrate over all the shells in the ball?
 
  • #3
Just evaluate the integral. Or is there something specific about the integral you don't understand?
 
  • #4
Yes I'm unsure of what my limits are in this case and what d[itex]\tau[/itex] is equal to
 
  • #5
The volume element ##d\tau## will depend on which coordinate system you choose to use. To calculate the potential at a point, you integrate over all space, but really you only need to worry about where the charge density ##\rho## doesn't vanish because everywhere else won't contribute to the integral.
 
  • #6
So would the integral from 0 to R work if d[itex]\tau[/itex] = r2sinθdrdd[itex]\phi[/itex] ?
 
  • #7
Is V = ρ0R4 / 4ε0 the answer to this problem?
 
  • #8
V = [itex]\frac{1}{4\pi\epsilon_{0}}[/itex][itex]∫^{2\pi}_{0}[/itex]d[itex]\phi[/itex][itex]∫^{\pi}_{0}[/itex]sin[itex]\theta[/itex]d[itex]\theta[/itex][itex]∫^{R}_{0}[/itex] [itex]\frac{ρ_{0}r^{2}}{r}[/itex] r[itex]^{2}[/itex]dr

Solving this:

V= [itex]\frac{ρ_{0}R^{4}}{4\epsilon_{0}}[/itex]
I'm quite happy with this answer, but if it is incorrect please let me know.
Thanks for the help
 
  • #9
I haven't worked it out, but your method looks fine. Does it match the answer you got using the previous method?
 

1. How do you define the charge density of a solid sphere?

The charge density of a solid sphere is defined as the amount of charge per unit volume of the sphere. It is represented by the Greek letter rho (ρ) and is calculated by dividing the total charge of the sphere by its volume.

2. What is the equation for finding the potential of a charged, solid sphere?

The equation for finding the potential of a charged, solid sphere is V = kQ/R, where V is the potential, k is the Coulomb constant, Q is the charge of the sphere, and R is the radius of the sphere.

3. How is the potential of a charged, solid sphere related to its charge density?

The potential of a charged, solid sphere is directly proportional to its charge density. This means that as the charge density increases, the potential also increases. This relationship is represented by the equation V ∝ ρ.

4. Can the potential of a charged, solid sphere be negative?

Yes, the potential of a charged, solid sphere can be negative. This occurs when the sphere has a net negative charge, which results in a negative potential. However, the magnitude of the potential is always positive.

5. How is the potential of a charged, solid sphere affected by changes in its charge density?

The potential of a charged, solid sphere is directly affected by changes in its charge density. An increase in charge density will result in a higher potential, while a decrease in charge density will result in a lower potential. This relationship is represented by the equation V ∝ ρ.

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