View Full Version : Just need a check...Finding total charge when given distribution
xxbigelxx
Sep15-10, 05:11 PM
1. The problem statement, all variables and given/known data
We have a sphere (radius R) where charge is placed inside of it, ρ(r)=ρ0 cos(θ/3) sin(φ/2). Find the total charge. Your answer will be in terms of R and p0.
2. Relevant equations
3. The attempt at a solution
I think I did it correctly, but I would just like to make sure I did (and thus ensure my understanding). My answer is marked by the star, and everything below it is just my integration work that was required by the problem.
gabbagabbahey
Sep15-10, 06:47 PM
1. The problem statement, all variables and given/known data
We have a sphere (radius R) where charge is placed inside of it, ρ(r)=ρ0 cos(θ/3) sin(φ/2). Find the total charge. Your answer will be in terms of R and p0.
2. Relevant equations
3. The attempt at a solution
I think I did it correctly, but I would just like to make sure I did (and thus ensure my understanding). My answer is marked by the star, and everything below it is just my integration work that was required by the problem.
I think you'll want to double check your integration of
\int_0^{\pi}\cos\left(\frac{\theta}{3}\right)\sin\ theta d\theta
xxbigelxx
Sep15-10, 07:24 PM
Hmm I tried to go over it again and I got the same result. Was it somewhere on the first integration by parts where I messed up? Thanks
xxbigelxx
Sep15-10, 07:47 PM
Ohhh I was double checking the wrong part. I ended up making a mistake at the very bottom. It changed my result from that integration to 27/16 and my final answer is now...
(Rho)(R^3)(9/4) C
I hope this should be correct now....
gabbagabbahey
Sep15-10, 08:00 PM
Ohhh I was double checking the wrong part. I ended up making a mistake at the very bottom. It changed my result from that integration to 27/16 and my final answer is now...
(Rho)(R^3)(9/4) C
I hope this should be correct now....
Looks good to me.
For future reference, you could have avoided IBP by using a trig identity:
\sin\theta\cos\left(\frac{\theta}{3}\right)=\frac{ \sin\left(\frac{4\theta}{3}\right)+\sin\left(\frac {2\theta}{3}\right)}{2}
xxbigelxx
Sep15-10, 08:23 PM
Ohh ok thanks. I don't even think I have ever come across that identity before.
gabbagabbahey
Sep16-10, 10:07 AM
Ohh ok thanks. I don't even think I have ever come across that identity before.
Its just an application of the identity
\sin\theta\cos\left(\phi\right)=\frac{ \sin\left(\theta+\phi)+\sin\left(\theta-\phi\right)}{2}
xxbigelxx
Sep16-10, 10:17 AM
Oh ok thanks a lot.
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