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Electric Potential inside insulating solid sphere. 
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#1
Sep2510, 10:11 PM

P: 63

1. The problem statement, all variables and given/known data
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. 2. Relevant 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: This is also what I found on different sources so I'm pretty sure the result is correct. 3. 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: 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. 


#2
Sep2510, 10:59 PM

HW Helper
P: 2,155

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
Sep2610, 12:26 AM

P: 63

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
Sep2610, 06:58 PM

P: 63

Electric Potential inside insulating solid sphere.
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
Sep2610, 08:42 PM

HW Helper
P: 2,155

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 suchandsuch 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
Sep2610, 11:53 PM

P: 63

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. 


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