Insulating Spherical Shell Potential Problem question

[ramuramu]

Homework Statement

I have a question about an upper-division E&M HW
problem I'm stuck on. Any help would be GREATLY APPRECIATED!

Problem Statement:

An insulating spherical shell of radius R was set up to have a potential
on its surface of V=A*cos^2(theta), where A is a constant. The potential
at a distance very far away from the shell is defined to be zero.

(a) Find the potential V(r,theta) both inside and outside the shell.

Homework Equations

E=-(grad)V

The Attempt at a Solution

Since E=-(grad)V , I took the gradient in spherical coordinates of the V expression above
and got

E={[(2*A)/R]cos(theta)sin(theta)}theta-hat


for the E field on the surface of the shell. I then let R become the
variable r and tried to do the line integral from "infinity" to R
to get the potential outside the shell. But this involves a dot product with dr
which is orthogonal to theta-hat so I got ZERO! which certainly
doesn't make sense...I would REALLY appreciate any advice someone might have on
this problem!

 

[fantispug]

Think carefully about what you're doing here.

Here's how I have read the question: You have a charged insulating spherical shell and no other sources of charge. You know the potential on the shell; deduce the potential at all other points.

Now the equation E=-grad(V) tells you how to get the electric field from the potential. I couldn't follow exactly what you were doing but it seemed like you calculated the electric field along the surface of a shell (you don't know what it is perpendicular to the shell yet) trying to add together infinitely many spherical shells to find the electric field of a solid sphere of infinite radius?

I think you need to consider general techniques for obtaining the potential due to a charged spherical shell, knowing the potential on the surface of the shell.

 

[ramuramu]

THANK YOU very much for your response.

Yes, I agree. I think I need to solve Laplace's Equation in spherical subject to the
boundary conditions:

(i) Vin = Vout
(ii) (dVin/dr - dVout/dr)|r=R = -(surface charge density)/epsilon-knot
(iii) As r-> infinity , V-> 0

Does this sound right to you?

Thank you so much again in advance!

 

[ramuramu]

I solved the problem using Laplace's Equation - the only problem is the Dielectic aspect of the shell. I suppose my boundary condition (ii) above:

(ii) (dVin/dr - dVout/dr)|r=R = -(surface charge density)/epsilon-knot

is wrong? Should I just change the epsilon-knot to the epsilon of the dielectric surface?
Although that doesn't seem to make too much sense to me. If there's no dielectric
VOLUME then does it not "matter" and I should just use epsilon-knot for the vacuum around?

Anyway, thank you so much again for your help! I REALLY appreciate it!

 

[paranormal]

Hello,

Thank you for reaching out for help with your E&M homework problem. It sounds like you have made a good start in trying to solve the problem by using the gradient of the potential function. However, as you mentioned, there seems to be an issue with the integration when converting from spherical coordinates to Cartesian coordinates.

One suggestion would be to try using the divergence theorem to solve this problem. The divergence theorem states that the flux through a closed surface is equal to the volume integral of the divergence of the vector field within the surface. In this case, the closed surface would be the spherical shell and the vector field would be the electric field. By setting up the appropriate integral using the divergence theorem, you should be able to solve for the potential both inside and outside the shell.

Another approach could be to use the method of images, which is a common technique used in electrostatics problems involving conductors and insulators. This method involves creating a mirrored charge distribution in order to satisfy boundary conditions and solve for the potential. There are many resources available online that can provide more information and examples of how to use this method.

I hope this helps and good luck with your homework! Remember, it's always a good idea to consult with your classmates and instructor if you are stuck on a problem. Keep up the good work!