New Reply

Conductors with off-centered cavity

 
Share Thread Thread Tools
Feb19-11, 10:57 PM   #1
 

Conductors with off-centered cavity


Suppose we had an uncharged solid spherical conductor with radius R, and some spherical cavity of radius a that is located b units above R, where a+b<R. If we place a charge inside the cavity at the center, how would you quantitatively express the E-field lines outside of the conductor?
PhysOrg.com
PhysOrg
physics news on PhysOrg.com

>> Study provides better understanding of water's freezing behavior at nanoscale
>> Soft matter offers new ways to study how ordered materials arrange themselves
>> Making quantum encryption practical
Feb19-11, 11:08 PM   #2
 
This is NOT a homework. Concentric cavities are too easy. I'm wondering if we can somehow apply uniqueness here to show that it actually doesn't matter where the cavity is...
Feb20-11, 12:11 AM   #3
 
The E field is easy to find using Gauss's law . And it doesn't matter where the charge is inside the sphere or the shape of the cavity.
Feb20-11, 12:37 AM   #4
 

Conductors with off-centered cavity


Quote by cragar View Post
The E field is easy to find using Gauss's law . And it doesn't matter where the charge is inside the sphere or the shape of the cavity.
Exactly. You can kind of reason it out in a way, by knowing that charges like to stay on the very outside of a conductor. Think about it, say you've got a positive charge inside that off-centre cavity. That charge is going to pull negative charges towards the inner surface of that cavity to balance it out. Then, to balance out those negative charges in the conductor, positive charges are going to show up on the outside of the sphere (regardless of where the cavity is), and they'll create an electric field outside the sphere exactly the same as if you had just a charged sphere.

You can use Gauss' Law to prove every part of that by setting up surfaces just outside the cavity and just outside the sphere and using the relation:

[tex]\Phi_E = \int\vec{E}\cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}[/tex]

Where you know [tex]\vec{E} = 0[/tex] inside the conductor.
Feb20-11, 01:07 AM   #5
 
True. But to solve that surface integral simply one must apply a symmetry argument. How do you arrive about that symmetry? The induced charge will not be evenly distributed on the outside of the sphere, but is there a way to apply uniqueness to arguing that?
Feb20-11, 06:02 AM   #6
 
Recognitions:
Homework Helper Homework Help
Science Advisor Science Advisor
The outer surface of the conductor will be an equipotential surface of radius R. An outside potential phi=Q/r satisfies this boundary condition. It is thus the unique potential outside the sphere.
It doesn't matter what goes on in cavities within the conduciting sphere.
New Reply
Thread Tools


Similar Threads for: Conductors with off-centered cavity
Thread Forum Replies
Origin of bar centered galaxy? Astrophysics 2
Configuration centered at the origin Differential Geometry 2
Non centered Parallax Question General Astronomy 0
centered projectile motion Introductory Physics Homework 1
f(x)=sin x, centered at x=pi/2 Question Differential Equations 8