Electric Potential inside insulating solid sphere.

  • Thread starter erogard
  • Start date
  • #1
62
0

Homework Statement



Hi everyone,
I'm supposed to find an expression for the electric potential as a function of r, the radial distance inside a solid and non conducting sphere of radius R. A total charge of q is uniformly distributed throughout its volume. The annoying part is that I'm supposed to do that using a relation between the potential and [tex] \rho[/tex] , the charge density.


Homework Equations



The equation the problem wants me to use is:

[tex] V( \textbf{r} ) = \frac{1}{4 \pi \epsilon_{0}} \int \frac{\rho ( \textbf{r} )}{r} d \tau ' [/tex]

[tex]r[/tex] being the distance between source & field (observation) point.

I already have obtained a result using a different method (integrating the electric field) and got:
21381469099254_picture1.jpg


This is also what I found on different sources so I'm pretty sure the result is correct.


The Attempt at a Solution



Yet I can't set up my integration is such a way so as to retrieve this result.
Using the law of cosines on the following figure:
5594471776872_untitled.jpg


I have: [tex] r^{2} = r'^{2} + z^{2} - 2 r' z cos ( \theta )[/tex]

Using spherical coordinates, I make the integral range from 0 to 2pi and 0 to pi for phi and theta.

For [tex]r'[/tex] I am more doubtful. Indeed, when we integrate with respect to the two angles first, we obtain an expression that involves the square root of the following difference: [tex](r' - z)^{2}[/tex].
Thus depending on where the observation point is on the sphere, the positive or negative root is to be taken. So I would guess I could split the integration into two parts, from R (the radius of the sphere) to r, and then from r to 0. Since [tex]\rho[/tex] is zero everywhere outside the sphere, there is no need to integrate from infinity, where the potential is assumed to go to zero.

Yet I do not get the expected result, and this using different integration limits. I am assuming I am doing something wrong initially, and if someone has any suggestion I would be very grateful for that.
 
Last edited:

Answers and Replies

  • #2
diazona
Homework Helper
2,175
8
For [tex]r'[/tex] I am more doubtful. Indeed, when we integrate with respect to the two angles first, we obtain an expression that involves the square root of the following difference: [tex](r' - z)^{2}[/tex].
In that element [tex](r' - z)^{2}[/tex], [tex]r'[/tex] and [tex]z[/tex] are simple distances, not vectors, right?
Thus depending on where the observation point is on the sphere, the positive or negative root is to be taken. So I would guess I could split the integration into two parts, from R (the radius of the sphere) to r, and then from r to 0. Since [tex]\rho[/tex] is zero everywhere outside the sphere, there is no need to integrate from infinity, where the potential is assumed to go to zero.
Don't you mean from R to z, then z to 0? (or rather, 0 to z and then z to R, I guess)

Subject to the couple of notes above, your setup seems fine. I did the problem setting it up the same way you've described, and I got the expected answer, so perhaps you have a math error somewhere. If you post the details of your work we can help you identify it.
 
  • #3
62
0
In that element [tex](r' - z)^{2}[/tex], [tex]r'[/tex] and [tex]z[/tex] are simple distances, not vectors, right?

Don't you mean from R to z, then z to 0? (or rather, 0 to z and then z to R, I guess)

Subject to the couple of notes above, your setup seems fine. I did the problem setting it up the same way you've described, and I got the expected answer, so perhaps you have a math error somewhere. If you post the details of your work we can help you identify it.

Hi diazona,

yes just distances, and yes again for z instead of r, that's what I meant indeed.

I'm going to post here the details of my calculations by Sunday afternoon, thanks a lot for taking a look at my problem.
 
  • #4
62
0
hi again,

I miraculously realized that I had a math mistake indeed when I was carrying my integration.
I obtained the exact same result, except that I got a minus sign in front of it.
You mentioned that the integration limits should be from 0 to z, then z to R. I did the opposite, which thus gave me the negative of my expected result - yet I'm a bit confused.

Isn't the integration assuming that our reference point is at infinity, thus we're "coming from" infinity, integrating from +∞ to z (or simply R to z since the charge density is zero outside the sphere) and then z to 0. That's what sounds logical to me even though it obviously leads the wrong answer. Do you know why?
 
  • #5
diazona
Homework Helper
2,175
8
You're mixing up two completely different equations. The one that requires a reference point at infinity is the definition of V as the path integral of E. When you calculate
[tex]\int_A^B\vec{E}\cdot\mathrm{d}\vec{s} = V(A) - V(B)[/tex]
all you really get is a difference - that is, you only know that the potential at point A is such-and-such relative to the potential at point B. In order to define "the potential at A" you have to have a standard reference point B.

In this problem, though, the integral defines V as the sum of contributions from each of the individual charge elements that make up the sphere. Conceptually you could think of it as a limit of
[tex]V = \sum_i V_i = \sum_i \frac{kq_i}{r_i}[/tex]
as the individual charge elements [itex]q_i[/itex] go to zero. There is no path involved here, and what you get isn't a difference between two values, so there's no need to have a standard reference point to compare everything to. The result you get is already absolute.
 
  • #6
62
0
You're mixing up two completely different equations. The one that requires a reference point at infinity is the definition of V as the path integral of E. When you calculate
[tex]\int_A^B\vec{E}\cdot\mathrm{d}\vec{s} = V(A) - V(B)[/tex]
all you really get is a difference - that is, you only know that the potential at point A is such-and-such relative to the potential at point B. In order to define "the potential at A" you have to have a standard reference point B.

In this problem, though, the integral defines V as the sum of contributions from each of the individual charge elements that make up the sphere. Conceptually you could think of it as a limit of
[tex]V = \sum_i V_i = \sum_i \frac{kq_i}{r_i}[/tex]
as the individual charge elements [itex]q_i[/itex] go to zero. There is no path involved here, and what you get isn't a difference between two values, so there's no need to have a standard reference point to compare everything to. The result you get is already absolute.


Oh goodness, I got it. In the second equation we're simply integrating the charge density over the whole sphere while there is no such idea of reference point I was getting myself confused with - we're just adding the "potential" contribution of each individual charge point. Thanks a lot for your help, I really appreciate it.
 

Related Threads on Electric Potential inside insulating solid sphere.

Replies
11
Views
918
Replies
3
Views
27K
Replies
2
Views
6K
Replies
1
Views
4K
Replies
1
Views
5K
Replies
1
Views
6K
  • Last Post
Replies
5
Views
9K
Replies
6
Views
3K
Replies
5
Views
4K
Replies
1
Views
3K
Top