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Grounded conductive shell

  • Thread starter Kosta1234
  • Start date
Problem Statement
Grounded conductive shell
Relevant Equations
image method
243780

Hello. I will be glad if someone can help me with this:

I've a grounded conductive shell with outer radius of ## R_2## and inner radius ##R_1##.
a charge ## Q ## is located inside of the shell, in distance ## r<R_1 ## , and a charge ## q ## is located in distance ## a > R_2 ## outside of the conductive shell.
The two charges are not located on the same axis and we don't know the angle between them.

What is the total charge on the shell?

My way of thinking is like this:
To split the problem to two main parts: the first one is a charge Q and radius ## R_1 ## shell. and the second is charge q and radius ## R_2 ## shell. I wanted to solve the laplace equation for each case - to figure out which image charges I should put so they would solve laplace equation and then, to solve this I would use: ## Q + q + Q_{image} = Q + q +Q_{shell} ##

to figure out the answer to the first part I used the method of image charges. I've ignored the fact that this is a shell with some given width, and found a potential that solved laplace's equation if the problem would be a charge Q and a shell with radius ## R_1 ##.
In this case I figured out that the image charge, which is located on the same axis that connecting charge Q with point (0,0,0) is:
## Q' = -Q \frac {R_1}{r} ##

the same thing with the second part:
## q' = -q \frac {R_2}{a} ##

so
## Q_{image} = -Q \frac {R_1}{r} -q \frac {R_2}{a} ##

it means that
## Q_{shell} = -Q \frac {R_1}{r} -q \frac {R_2}{a} ##



But I saw the answer, and she is:
## Q_{shell } =
-Q -q \frac {R_2}{a} ## .

Where I've been wrong?

Thank you.
 
Last edited:

PeroK

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If you split the problem into a superposition, then the solution on the inner surface doesn't need the image method.
 
why is that?
 
Because the net flux is zero?
Using Gauss law is not only when I got symmetry in the problem?
 

PeroK

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Because the net flux is zero?
Using Gauss law is not only when I got symmetry in the problem?
If the enclosed charge is not zero, then the field can't be zero everywhere inside the conductor. Your solution cannot, therefore, be valid for the inner surface.

The converse is not true. If the enclosed charge is zero, the total flux through a surface is zero, but the field may not be zero everywhere. That's why you need a specific charge distribution on the inner surface to have a zero field externally.

But the total charge on the inner surface must be equal and opposite to the point charge ##Q##.
 
Thank you!.
 

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