- #1
davidbenari
- 466
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My question is best illustrated by an example from a Griffiths book on E&M:
"A point charge q is situated a distance ##a## from the center of a grounded conducting sphere of radius R (##a>R##). Find the potential outside the sphere... With the addition of a second charge you can simulate any ##V_0## on the sphere now... Solve this problem now with a neutral conducting sphere at potential ##V_0##."
Now, this problem is pretty well-known and it can be shown that adding a charge ##q'=-\frac{R}{a}q## a distance ##b=\frac{R^2}{a}## collinear with the origin of the sphere and the original charge ##q##, solves the boundary problem for a grounded sphere. If I want to create a new boundary voltage ##V_0## it is simple to recognize that I have to put a second image charge ##q'' ## on the origin such that ##V_0=\frac{kq''}{R}##.
But I'm confused as to why for a neutral conducting sphere my two image charges have to satisfy ##q''+q'=0##. Uniqueness theorems talk about boundary conditions, and in a way I only have to satisfy a specified voltage ##V_0## on the boundaries, why do I also have to satisfy the conditions of charge inside this sphere? What part of the uniqueness theorem say so?
I know the uniqueness theorems talk about specifying voltage or net charge on a boundary. But when solving a PDE giving both net charge and voltage on a conducting surface usually overdetermines the system, and talking about net charge gives you information about a surface integral which I think is pretty useless information due to the complexity of such a condition.
For anyone not acquainted with this problem, I think they state it clearly on this website:http://ocw.nctu.edu.tw/upload/classbfs120903470749199.pdf
"A point charge q is situated a distance ##a## from the center of a grounded conducting sphere of radius R (##a>R##). Find the potential outside the sphere... With the addition of a second charge you can simulate any ##V_0## on the sphere now... Solve this problem now with a neutral conducting sphere at potential ##V_0##."
Now, this problem is pretty well-known and it can be shown that adding a charge ##q'=-\frac{R}{a}q## a distance ##b=\frac{R^2}{a}## collinear with the origin of the sphere and the original charge ##q##, solves the boundary problem for a grounded sphere. If I want to create a new boundary voltage ##V_0## it is simple to recognize that I have to put a second image charge ##q'' ## on the origin such that ##V_0=\frac{kq''}{R}##.
But I'm confused as to why for a neutral conducting sphere my two image charges have to satisfy ##q''+q'=0##. Uniqueness theorems talk about boundary conditions, and in a way I only have to satisfy a specified voltage ##V_0## on the boundaries, why do I also have to satisfy the conditions of charge inside this sphere? What part of the uniqueness theorem say so?
I know the uniqueness theorems talk about specifying voltage or net charge on a boundary. But when solving a PDE giving both net charge and voltage on a conducting surface usually overdetermines the system, and talking about net charge gives you information about a surface integral which I think is pretty useless information due to the complexity of such a condition.
For anyone not acquainted with this problem, I think they state it clearly on this website:http://ocw.nctu.edu.tw/upload/classbfs120903470749199.pdf
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