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Potential inside of chrged nonconductive sphere 
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#1
Oct308, 08:09 AM

P: 11

1. The problem statement, all variables and given/known data
A charge is distributed uniformly throughout a nonconducting spherical volume of radius R. Show, that the potential at distance r from center (r<R) is given by: V=q(3*R^2r^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? 


#2
Oct308, 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) 


#3
Oct308, 08:13 AM

P: 328

And R = radius of gaussian surface
r = radius of your sphere? please be precise. 


#4
Oct308, 08:26 AM

P: 328

Potential inside of chrged nonconductive 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^2R^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.. 


#5
Oct308, 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? 


#6
Oct308, 08:56 AM

P: 11

Sorry again: from inf to r not from r to inf, I meant.



#7
Oct308, 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... 


#8
Oct308, 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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