Sketching the Potential of a Spherical Shell with Uniformly Distributed Charge

In summary, the conversation discusses finding the potential at different points within a spherical shell with a uniform charge distribution. The potential at the center, inner radius, and outer radius are all solved using different methods such as Gauss's Law and integrals. The equation P = k * Q/((R+r)/2) is mentioned as a way to solve for the potential at the center and inner radius, but its derivation is not fully understood. The topic of electric fields and conductors is also brought up.
  • #1
singinglupine
15
0
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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  • #2
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
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
singinglupine said:
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.
 
  • #5
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
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
Team_Zulu said:
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?
 
  • #8
zachzach said:
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.
 
  • #9
Team_Zulu said:
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.
 
  • #10
zachzach said:
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.
 

1. What is the potential of a spherical shell?

The potential of a spherical shell is the amount of energy required to bring a unit test charge from infinity to a point on the surface of the shell.

2. How is the potential of a spherical shell calculated?

The potential of a spherical shell can be calculated using the formula V = kQ/r, where k is the Coulomb constant, Q is the charge of the shell, and r is the distance from the center of the shell to the point where the potential is being measured.

3. Does the potential of a spherical shell depend on the distance from the center?

No, the potential of a spherical shell is the same at all points on the surface of the shell, regardless of the distance from the center.

4. What is the relationship between the potential of a spherical shell and its electric field?

The electric field inside a spherical shell is zero, so the potential is constant at all points inside the shell. However, the potential varies outside the shell and is directly proportional to the electric field at those points.

5. Can the potential of a spherical shell be negative?

Yes, the potential of a spherical shell can be negative if the charge of the shell is negative. This indicates that it requires energy to bring a positive test charge from infinity to a point on the surface of the shell, and the potential decreases as the distance from the shell increases.

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