Spherical Shells - Potential


by singinglupine
Tags: potential, shells, spherical
singinglupine
singinglupine is offline
#1
Apr21-10, 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 10-8 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?
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zachzach
zachzach is offline
#2
Apr21-10, 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?
singinglupine
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#3
Apr21-10, 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.

Team_Zulu
Team_Zulu is offline
#4
Apr23-10, 05:17 AM
P: 3

Spherical Shells - Potential


Quote Quote by singinglupine View Post
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.
Would you care to elaborate on this? I am trying to solve this same type of problem and cannot figure it out. I have searched through the book and on the internet for the past 6 hours and have come up with nothing.

It'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.
singinglupine
singinglupine is offline
#5
Apr23-10, 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?
ehild
ehild is offline
#6
Apr23-10, 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
zachzach
zachzach is offline
#7
Apr23-10, 03:21 PM
P: 258
Quote Quote by Team_Zulu View Post
Would you care to elaborate on this? I am trying to solve this same type of problem and cannot figure it out. I have searched through the book and on the internet for the past 6 hours and have come up with nothing.

It'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.
There is 3 different values for the electric field. Inside the inner shell, in between the shells, and outside the outer shell. They can all be solved by Gauss's law and you do not need to use a triple integral. Draw your Gaussian sphere at an arbitrary r in between the two shells. What does integral simplify too? What is the enclosed charge? Remember p = Q/V --> Q = pV. What is your volume though?
Team_Zulu
Team_Zulu is offline
#8
Apr27-10, 07:40 PM
P: 3
Quote Quote by zachzach View Post
There is 3 different values for the electric field. Inside the inner shell, in between the shells, and outside the outer shell. They can all be solved by Gauss's law and you do not need to use a triple integral. Draw your Gaussian sphere at an arbitrary r in between the two shells. What does integral simplify too? What is the enclosed charge? Remember p = Q/V --> Q = pV. What is your volume though?
There is only one shell. The electric field inside is zero. The electric field in the conducting shell is zero and the electric field outside of the shell is E = k*Q/r^2. I really don't see how any of that helps you solve this problem. My friend told me the equation to solve it, which is P = k * Q/((R+r)/2)

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.
zachzach
zachzach is offline
#9
Apr27-10, 07:47 PM
P: 258
Quote Quote by Team_Zulu View Post
There is only one shell. The electric field inside is zero. The electric field in the conducting shell is zero and the electric field outside of the shell is E = k*Q/r^2. I really don't see how any of that helps you solve this problem. My friend told me the equation to solve it, which is P = k * Q/((R+r)/2)

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.
How is it a conducting shell when the charge is uniformly distributed over its volume? If it was a conductor all charges would be on the surface so it would be uniformly distributed over its area.
Team_Zulu
Team_Zulu is offline
#10
Apr27-10, 07:59 PM
P: 3
Quote Quote by zachzach View Post
How is it a conducting shell when the charge is uniformly distributed over its volume? If it was a conductor all charges would be on the surface so it would be uniformly distributed over its area.
You are correct. It's not a conducting shell. Thanks for pointing that out. It makes a lot more sense now that I realize that.

Thanks again.


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