Visualizing the Potential of a Spherical Shell

AI Thread Summary
The discussion focuses on calculating the electric potential of a spherical shell with an inner radius of 14.6 cm and an outer radius of 15.2 cm, carrying a uniform charge of 5.35 × 10^-8 C. Participants explore the shape of the potential graph, noting positive asymptotes at the outer radius and questioning the behavior within the inner radius. It is clarified that the potential at the center of the shell is the same as at any point inside the inner radius due to the uniform charge distribution. The integration method for determining potential involves using the relationship between electric field and potential, emphasizing that the electric field inside the shell is zero, leading to constant potential. The conversation concludes with a better understanding of the potential's behavior within and outside the shell.
shoe02
Messages
4
Reaction score
0

Homework Statement


The inner radius of a spherical shell is 14.6 cm, and the outer radius is 15.2 cm. The shell carries a charge of 5.35 × 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 would the graph look like? i think it has positive asymptotes at r = +/- 15.2cm (assuming r = 0 is at the origin) but I am not whether it crosses the axis on the interval [-14.6, 14.6] or whether it comes to a minimum
What is the potential at the center of the shell (r=0)?

im not sure how to approach this problem...

-thanks
 
Physics news on Phys.org
By way of a hint, the problem is nearly identical to Newtonian gravity for a spherically symmetric body of uniform mass density.
 
Last edited:
should the potential be zero due to the equation V = Kq/r ?
 
shoe02 said:
should the potential be zero due to the equation V = Kq/r ?

Nope.

You mist know that the potential at any point is the amount of work done in bringing a charge from infinity to that point.

The work done to bring from infinity to outer surface can be found from your formula given above.

In the region bound by the inner and the outer surfaces you will have to find the potential by integrating infinetisally small shells whose radii range from the given inner radii to the outer. As far as the potential at the center it would be same as that as any other point inside the inner radii
 
ok thanks. that helps a lot, and i didnt think of potential that way, but it makes a lot more sense now.

thanks again for the help
 
well so...why would we be the potential be same for the points inside the inner radii. And how will you integrate? Hint E.dx = dV
 
is the equation i integrate this: ((K*q)/r^2)dr
and is it the potential same throughout the shell because the field inside the shell is linear?

again, thanks for the help
 
Last edited:
Back
Top