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

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SUMMARY

The discussion focuses on calculating the electric potential at the center of a solid sphere with a volume charge density defined as ρ(r) = ρ0r². The potential is derived using the formula V = (1/4∏ε)∫(ρ/r)dτ, where the integration is performed over the volume of the sphere. The correct approach involves integrating from 0 to R, with the volume element dτ expressed in spherical coordinates as dτ = r²sinθdrdφ. The final result for the potential at the origin is confirmed as V = ρ0R⁴ / 4ε0.

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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\tau


The Attempt at a Solution



I don't need the full solution, just the first step would be much appreciated.
 
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Find the potential at the center due to a thin shell of charge and then integrate over all the shells in the ball?
 
Just evaluate the integral. Or is there something specific about the integral you don't understand?
 
Yes I'm unsure of what my limits are in this case and what d\tau is equal to
 
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.
 
So would the integral from 0 to R work if d\tau = r2sinθdrdd\phi ?
 
Is V = ρ0R4 / 4ε0 the answer to this problem?
 
V = \frac{1}{4\pi\epsilon_{0}}∫^{2\pi}_{0}d\phi∫^{\pi}_{0}sin\thetad\theta∫^{R}_{0} \frac{ρ_{0}r^{2}}{r} r^{2}dr

Solving this:

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

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