# Homework Help: Finding voltage due to spherical charge

1. Jul 26, 2013

### eminem14

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.

Last edited: Jul 26, 2013
2. Jul 26, 2013

### haruspex

What do you know about the potential due to a uniform spherical shell?

3. Jul 26, 2013

### eminem14

it's equal to [1/(4*pi*epsilon)] *Q / r

4. Jul 27, 2013

### haruspex

Where r is what?
Then think of the given charge density as a set of such shells.

5. Jul 27, 2013

### 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?

6. Jul 27, 2013

### haruspex

As long as the point for which you want the potential is outside all of the shells, the radius of each shell becomes irrelevant.

7. Jul 27, 2013

### eminem14

so that would mean that the r - terms in the integral cancel ? and if radii are irrelevant then what is relevant besides θ & ∅?

8. Jul 27, 2013

### 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?

9. Jul 27, 2013

### 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. 1<r<1.5 2. 0 <= θ <= ∏ 3. 0 <= ∅ <= 2∏

10. Jul 27, 2013

### 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<r<1.5 0 <= θ <= ∏ 0 <= ∅ <= 2∏

*the exponent of the denominator is 3/2 not 3

Last edited: Jul 27, 2013
11. Jul 27, 2013

### 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.

12. Jul 27, 2013

### eminem14

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 ?

Last edited: Jul 27, 2013
13. Jul 28, 2013

### 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?

14. Jul 28, 2013

### eminem14

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

15. Jul 28, 2013

### 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?

16. Jul 28, 2013

### eminem14

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?

17. Jul 28, 2013

### haruspex

Yes.
It's the Q/r that's wrong. It should be Q/s.

18. Jul 28, 2013

### eminem14

S is a constant though right?

19. Jul 28, 2013

### haruspex

For a given reference point, yes.

20. Jul 28, 2013

### eminem14

Thanks for the help got the correct result. But I just wanna understand something, why is it that the radii are irrelevant?

21. Jul 28, 2013

### haruspex

For a uniformly charged spherical shell, the field outside the shell does not depend on the radius of the shell. It only depends on the total charge and how far you are from the centre of the shell. There's no simple reason for this - it just turns out that way for inverse square laws operating in three dimensions. Similarly, there is no field inside the shell. So, in general, if you have a solid sphere in which the charge only depends on the distance from the centre, and you want the potential at some distance s from that centre, you can throw away all parts of the sphere further than s from the centre and consider all the charge that remains as being concentrated at the centre.
The same applies for gravitational fields. If you pretend that the earth is uniformly dense, this gives the result that the field inside the earth is proportional to distance from the centre.

22. Jul 28, 2013

### eminem14

Thanks again and one final dumb question are the units for the voltage volts or mili volts?

23. Jul 28, 2013

### haruspex

You were given distances in m and charge in μC. Had the charge been in Coulombs then the answer would have been in Volts, right?

24. Jul 29, 2013

### eminem14

Thats what i figured. In my calculations i changed the μC to C and assumed the voltage would be in volts. Hopefully that should be correct.