
#1
Apr2110, 02:20 PM

P: 15

The inner radius of a spherical shell is 14.6 cm, and the outer radius is 15.0 cm. The shell carries a charge of 5.85 × 108 C, distributed uniformly though its volume. Sketch, for your own benefit, the graph of the potential for all values of r (the radial distance from the center of the shell).
What is the potential at the center of the shell (r=0)? What is the potential at the inner radius? What is the potential at the outer radius?  I solved this using V = kQ/r. This part makes sense to me. The first two questions should have the same answer I know. I'm not sure how to solve for them, since I know I can't use the same equation that I did for the outer ring. Is there a way without full out integrating? 



#2
Apr2110, 03:59 PM

P: 258

Try Gauss's Law for in between the inner and outer radii and for inside both radii. Then, what do you know about V at a boundary?




#3
Apr2110, 08:23 PM

P: 15

Thanks, I figured it out at a help session. It dealt with volume integrals, and the charge density per volume, something I've never learned in calc or physics.




#4
Apr2310, 05:17 AM

P: 3

Spherical Shells  PotentialIt's really frustration to get to the same point every time just to have the person say "I figured it out". I understand how to get the potential on the outer surface by integrating kQ/r^2 from infinity to R. I also understand that the electric field within the spherical shell is zero and that this causes the potential within the sphere to be constant, but this doesn't help me get an answer. It just tells me that the answer to part A is also the answer to part B. I just don't know where to go from here. 



#5
Apr2310, 07:45 AM

P: 15

Yeah, those are the same things I understood.
I'm sorry, I would explain it to you, but apparently since it involves triple integrals, it's much beyond my math level. I can give you the equations for roh, surface charge density per volume (like lambda for linear, sigma for area) and the final equation you end up deriving. Is this a problem on LON CAPA for you too? 



#6
Apr2310, 09:43 AM

HW Helper
Thanks
P: 9,818

Apply Gauss's law to get the electric field. Because of spherical symmetry, E has only radial component and depends only on r.
The potential is obtained by integration from r to infinity where the potential is taken 0. ehild 



#7
Apr2310, 03:21 PM

P: 258





#8
Apr2710, 07:40 PM

P: 3

I know that the electric field within the shell is zero so the electric potential is the same for the center of the sphere and inner radius, but I don't understand how the equation was derived. Logically I guess it makes sense, but I would still like to know how that equation was derived. 



#9
Apr2710, 07:47 PM

P: 258





#10
Apr2710, 07:59 PM

P: 3

Thanks again. 


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