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Potential inside of chrged non-conductive sphere

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anizet
#1
Oct3-08, 08:09 AM
P: 11
1. The problem statement, all variables and given/known data
A charge is distributed uniformly throughout a non-conducting spherical volume of radius R. Show, that the potential at distance r from center (r<R) is given by:
V=q(3*R^2-r^2)/(8*PI*e0*R^3)

2. Relevant equations
From Gauss low:
E(r)=q*r/(4*PI*e0*R^3)

E(r)=-grad(V(r))

3. The attempt at a solution
After integrating I got:
V(r)=-q*R^2/(8*PI*e0*R^3)
which is in fact stupid, as it yields V=0 at r=0. So probably I integrated with wrong limits.
How to obtain correct formula?
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anizet
#2
Oct3-08, 08:11 AM
P: 11
I made a small mistake in typing the formula on V that I obtained. It should be:
V(r)=-q*r^2/(8*PI*e0*R^3)
Rake-MC
#3
Oct3-08, 08:13 AM
P: 328
And R = radius of gaussian surface
r = radius of your sphere?
please be precise.

Rake-MC
#4
Oct3-08, 08:26 AM
P: 328
Potential inside of chrged non-conductive sphere

Never mind, I've solved it.

You have two different functions for the electric field.
If r > R, [tex] E = \frac{kQ}{r^2} [/tex] where [tex] k = \frac{1}{4\pi\epsilon_0} [/tex]
if r < R, [tex] E = \frac{kQr}{R^3} [/tex]

so V of r < R will be the negative integral of r > R from R to infinity plus the negative integral of r < R from r to R.

[tex] V = - \int^{R}_{\infty} \frac{kQ}{r^2}dr - \int^{r}_{R} \frac{kQr}{R^3}dr [/tex]

[tex] V = \frac{kR}{R} - \frac{kR}{2R^3}(r^2-R^2) = \frac{2kQR^2}{2R^3} + \frac{kQR^2}{2R^3} - \frac{kQr^2}{2R^3} = \frac{kQ(3R^2 - r^2)}{2R^3} [/tex]

subbing k back in = [tex] \frac{Q(3R^2 - r^2)}{8 \pi \epsilon_0 R^3} [/tex]

phew.. Hope there are no errors in that haha. I have been up for far too many consecutive hours..
anizet
#5
Oct3-08, 08:55 AM
P: 11
Thank you very much for the reply!
I have just one more question: why should I integrate from infinity to zero, not from zero to infinity?
anizet
#6
Oct3-08, 08:56 AM
P: 11
Sorry again: from inf to r not from r to inf, I meant.
anizet
#7
Oct3-08, 08:58 AM
P: 11
One more correction:
Why should I integrate from infinity to r not from zero to r?
I hope, now it's ok...
Rake-MC
#8
Oct3-08, 09:52 AM
P: 328
because if we integrated from zero to r, that would give us V for inside the gaussian surface aswell as inside the sphere.

Sorry for the late reply I was away for sometime.


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