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Old_sm0key
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SORRY TITLE SHOULD READ: "JUSTIFICATION FOR 0 NET E FIELD WITHIN A CHARGED SHELL"
<< Title fixed by Moderator (except for the all-caps of course) >>
1. Homework Statement
A conceptual query:
Considering an irregular shaped conducing shell with a net charge, using Gauss' Law, any Gaussian surface located within the shell will enclosed 0 charge. Hence the flux integral vanishes.
But clearly these charges are radiating field lines within the shell, so each elemental flux, [itex]\underline{E}.d\underline{A}[/itex], will be non zero...? Thus why is that the LHS Gauss' Law would be 0?
[itex]\displaystyle\int_S\underline{E}.d\underline{A} = \frac{Q}{\epsilon_0}[/itex]
(see part 1)
<< Title fixed by Moderator (except for the all-caps of course) >>
1. Homework Statement
A conceptual query:
Considering an irregular shaped conducing shell with a net charge, using Gauss' Law, any Gaussian surface located within the shell will enclosed 0 charge. Hence the flux integral vanishes.
But clearly these charges are radiating field lines within the shell, so each elemental flux, [itex]\underline{E}.d\underline{A}[/itex], will be non zero...? Thus why is that the LHS Gauss' Law would be 0?
Homework Equations
[itex]\displaystyle\int_S\underline{E}.d\underline{A} = \frac{Q}{\epsilon_0}[/itex]
The Attempt at a Solution
(see part 1)
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