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Finding voltage due to spherical charge

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eminem14
#1
Jul26-13, 09:06 PM
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<r<1.5m. Find the potential difference between point A(3.0,4.0,12.0) and point B(2.0,2.0,2.0)


2. Relevant equations
dQ=(ρv)dv dV=[dQ/(4∏ε0]*norm(R) where R is position vector of a point charge relative to observation point A or B.
dv=r^2 *sinθ*dr*dθ*d∅
x=rsinθ*cos∅ ; y=r*sinθ*sin∅ ; z=r*cosθ (transformations of coordiantes from spherical to cartesian
Ar= Ax*sinθ*cos∅ + Ay*sinθ*sin∅ + Az*cosθ (obtaining radial component(Ar) from cartesian components Ax,Ay and Az)
3. The attempt at a solution
I attempted to first find the voltage at A and B and subtract them from each other by doing the following:
1. Finding dQ=[1/(r^2)]*r^2 *sinθ*dr*dθ*d∅=sinθ*dr*dθ*d∅
2. Therefore, dV=sinθ*dr*dθ*d∅*norm(R)
3. Rx= r*sinθ*cos∅ -3 Ry=r*sinθ*sin∅ -4 Rz=r*cosθ -12 (trasformation of spherical components to cartesian coordinates minus the coordinates of A.

Problem:
I am finding difficulty in getting the norm of R because I can't obtain the unit vector of R which means I can't evaluate the integral to find V. I assumed that the voltage would only have a radial component given the method used in the matlab solution I have(not sure though and dont know why). Also given the subtractions shown in point 3, i couldnt integrate because the terms under the root for the magnitude of R can't be simplified.

Please help and thank u in advance
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haruspex
#2
Jul26-13, 11:09 PM
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What do you know about the potential due to a uniform spherical shell?
eminem14
#3
Jul26-13, 11:31 PM
P: 19
it's equal to [1/(4*pi*epsilon)] *Q / r

haruspex
#4
Jul27-13, 01:20 AM
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Finding voltage due to spherical charge

Quote Quote by eminem14 View Post
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.
eminem14
#5
Jul27-13, 02:22 AM
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?
haruspex
#6
Jul27-13, 02:31 AM
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Quote Quote by eminem14 View Post
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.
eminem14
#7
Jul27-13, 11:53 AM
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 & ∅?
haruspex
#8
Jul27-13, 04:47 PM
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Quote Quote by eminem14 View Post
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?
eminem14
#9
Jul27-13, 06:42 PM
P: 19
([10^-6]/(4*∏*ε)*∫∫∫[itex]\frac{sinθ*dr*dθ*d∅}{(r*sinθcos∅ - 3)^2 + (r*sinθ*sin∅ -4)^2 + (r*cosθ - 12)^2)^3/2}[/itex]

where the limits of the integral: 1. 1<r<1.5 2. 0 <= θ <= ∏ 3. 0 <= ∅ <= 2∏
eminem14
#10
Jul27-13, 06:47 PM
P: 19
Quote Quote by haruspex View Post
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)*∫∫∫[itex]\frac{sinθ*dr*dθ*d∅}{(r*sinθcos∅ - 3)^2 + (r*sinθ*sin∅ -4)^2 + (r*cosθ - 12)^2)^3/2}[/itex]
where the limits of the integral: 1<r<1.5 0 <= θ <= ∏ 0 <= ∅ <= 2∏

*the exponent of the denominator is 3/2 not 3
haruspex
#11
Jul27-13, 08:37 PM
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Quote Quote by eminem14 View Post
([10^-6]/(4*∏*ε)*∫∫∫[itex]\frac{sinθ*dr*dθ*d∅}{(r*sinθcos∅ - 3)^2 + (r*sinθ*sin∅ -4)^2 + (r*cosθ - 12)^2)^3/2}[/itex]
where the limits of the integral: 1<r<1.5 0 <= θ <= ∏ 0 <= ∅ <= 2∏

*the exponent of the denominator is 3/2 not 3
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.
eminem14
#12
Jul27-13, 09:22 PM
P: 19
Quote Quote by haruspex View Post
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 ?
haruspex
#13
Jul28-13, 01:37 AM
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Quote Quote by eminem14 View Post
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?
eminem14
#14
Jul28-13, 10:00 AM
P: 19
Quote Quote by haruspex View Post
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
haruspex
#15
Jul28-13, 04:40 PM
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Quote Quote by eminem14 View Post
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?
eminem14
#16
Jul28-13, 04:45 PM
P: 19
Quote Quote by haruspex View Post
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?
haruspex
#17
Jul28-13, 06:45 PM
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Quote Quote by eminem14 View Post
The min and max for the shells are 1.0 and 1.5. So those r the limits of the integral
Yes.
but then where does s fall into the integral. Is it a constant multiplied by the integral?
It's the Q/r that's wrong. It should be Q/s.
eminem14
#18
Jul28-13, 06:54 PM
P: 19
S is a constant though right?


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