# Finding voltage due to spherical charge

 P: 19 1. The problem statement, all variables and given/known data A volume charge density of ρv=1/r^2 μC/m^3 exists in the region bounded by 1.0m
 Homework Sci Advisor HW Helper Thanks P: 9,863 What do you know about the potential due to a uniform spherical shell?
 P: 19 it's equal to [1/(4*pi*epsilon)] *Q / r
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Finding voltage due to spherical charge

 Quote by eminem14 it's equal to [1/(4*pi*epsilon)] *Q / r
Where r is what?
Then think of the given charge density as a set of such shells.
 P: 19 r is the distance from the centre of the shell to the point charge you're trying to calculate the potential at. But i dont get how thinking of the charge density as a set of uniformly charged shells simplifies things?
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 Quote by eminem14 r is the distance from the centre of the shell to the point charge you're trying to calculate the potential at. But i dont get how thinking of the charge density as a set of uniformly charged shells simplifies things?
As long as the point for which you want the potential is outside all of the shells, the radius of each shell becomes irrelevant.
 P: 19 so that would mean that the r - terms in the integral cancel ? and if radii are irrelevant then what is relevant besides θ & ∅?
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 Quote by eminem14 so that would mean that the r - terms in the integral cancel ? and if radii are irrelevant then what is relevant besides θ & ∅?
You've been using r to mean two different things. One is relevant, the other isn't. I'll switch one of them to s.
Consider a shell radius r, thickness dr, and a point distance s > r from its centre. You correctly gave the formula for the potential at that point as [1/(4*pi*epsilon)] *Q / s. In your set-up, the charge density on each such shell is a function of r. Can you write down the integral?
 P: 19 ([10^-6]/(4*∏*ε)*∫∫∫$\frac{sinθ*dr*dθ*d∅}{(r*sinθcos∅ - 3)^2 + (r*sinθ*sin∅ -4)^2 + (r*cosθ - 12)^2)^3/2}$ where the limits of the integral: 1. 1
P: 19
 Quote by haruspex You've been using r to mean two different things. One is relevant, the other isn't. I'll switch one of them to s. Consider a shell radius r, thickness dr, and a point distance s > r from its centre. You correctly gave the formula for the potential at that point as [1/(4*pi*epsilon)] *Q / s. In your set-up, the charge density on each such shell is a function of r. Can you write down the integral?
([10^-6]/(4*∏*ε)*∫∫∫$\frac{sinθ*dr*dθ*d∅}{(r*sinθcos∅ - 3)^2 + (r*sinθ*sin∅ -4)^2 + (r*cosθ - 12)^2)^3/2}$
where the limits of the integral: 1<r<1.5 0 <= θ <= ∏ 0 <= ∅ <= 2∏

*the exponent of the denominator is 3/2 not 3
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 Quote by eminem14 ([10^-6]/(4*∏*ε)*∫∫∫$\frac{sinθ*dr*dθ*d∅}{(r*sinθcos∅ - 3)^2 + (r*sinθ*sin∅ -4)^2 + (r*cosθ - 12)^2)^3/2}$ where the limits of the integral: 1
You don't need any phis or thetas. You have a set of shells radius r, thickness dr. You know the volume of each shell and its charge density, so you know its total charge. You have a reference point distance s > max r from the centre of all these shells, so you know the potential at that point due to each shell. The potential due to the whole set is just the integral wrt r.
P: 19
 Quote by haruspex You don't need any phis or thetas. You have a set of shells radius r, thickness dr. You know the volume of each shell and its charge density, so you know its total charge. You have a reference point distance s > max r from the centre of all these shells, so you know the potential at that point due to each shell. The potential due to the whole set is just the integral wrt r.
so s>1.5
what would the integral be? isnt the volume of each shell would 4/3 * pi * (rout^3 - rin ^3)? and what about the coordinates of A & B? where do they fall into this
if Q can be calculated would the integral be:
k*Q*∫1/r^3 dr r between 1 and 1.5 ?
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 Quote by eminem14 so s>1.5
s is whatever it is for each of these two points in turn, but it looks clear that it will be > 1.5 for each of A(3.0,4.0,12.0), B(2.0,2.0,2.0)
 what would the integral be? isnt the volume of each shell would 4/3 * pi * (rout^3 - rin ^3)?
For a thin shell, you can take the volume as area * dr = 4πr2dr. You are given the charge density there as 1/r2 μC/vol. So what is the total charge on the shell?
P: 19
 Quote by haruspex s is whatever it is for each of these two points in turn, but it looks clear that it will be > 1.5 for each of A(3.0,4.0,12.0), B(2.0,2.0,2.0) For a thin shell, you can take the volume as area * dr = 4πr2dr. You are given the charge density there as 1/r2 μC/vol. So what is the total charge on the shell?
Q would be (1/r2 * 4πr2dr)= 4∏ dr

∴ ∫kQ/r= 1/ε∫1/r dr for point A limits of the integral: 1.5 < r < s

and ∫kQ/r= 10^-6/ε∫1/r dr for point B limits of the integral: 1.5 < r < s

but how is s calculated for each point? is it the magnitude of position vector pointing to the point → mag(3i + 4j + 12k ) = root(169)= 13 seems too big
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 Quote by eminem14 Q would be (1/r2 * 4πr2dr)= 4∏ dr
Yes.
 ∴ ∫kQ/r= 1/ε∫1/r dr
You're confusing the two distances again. Try to stick to the notation I introduced: r is for the radius of a shell; s is for the distance from the centre of the shell to the point of interest.
 for point A limits of the integral: 1.5 < r < s
Wrong limits. What are the min and max r for the shells?
 but how is s calculated for each point? is it the magnitude of position vector pointing to the point → mag(3i + 4j + 12k ) = root(169)= 13 seems too big
Yes, it's that. Why does it seem too big?
P: 19
 Quote by haruspex Yes. You're confusing the two distances again. Try to stick to the notation I introduced: r is for the radius of a shell; s is for the distance from the centre of the shell to the point of interest. Wrong limits. What are the min and max r for the shells? Yes, it's that. Why does it seem too big?
The min and max for the shells are 1.0 and 1.5. So those r the limits of the integral but then where does s fall into the integral. Is it a constant multiplied by the integral?
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