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## Homework Statement

## Homework Equations

## The Attempt at a Solution

The textbook says that the electric field on a surface of a conductor is:

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- Thread starter Jozefina Gramatikova
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- #1

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The textbook says that the electric field on a surface of a conductor is:

- #2

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- #3

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what do you mean by "if there is free space on both sides"?

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A layer of surface charge where you have air on both sides when you draw what is called a Gaussian pillbox around a section of area ## A ## of surface charge density ## \sigma ##. When you calculate the flux on the "inside", the conductor necessarily has ## E=0 ## inside the conductor, so there is no contribution to the flux from the inside surface. Alternatively, if there is air on both sides, both sides get the same outward pointing ## E ## and the result is ## E=\frac{\sigma}{2 \epsilon_o} ##. ## \\ ## (In full detail, for the conductor case: ## \int E \cdot dA=EA=\frac{\sigma A}{\epsilon_o} ##, so that ## E=\frac{\sigma}{\epsilon_o} ##. ## \\ ## And for the case with air on both sides: ## \int E \cdot dA=2 E A=\frac{\sigma A}{\epsilon_o } ##, so that ## E=\frac{\sigma}{2 \epsilon_o} ## ).what do you mean by "if there is free space on both sides"?

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- #5

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Oh, damn. I finally understood it! Thank you so much! It actually makes sense!A layer of surface charge where you have air on both sides when you draw what is called a Gaussian pillbox around a section of area ## A ## of surface charge density ## \sigma ##. When you calculate the flux on the "inside", the conductor necessarily has ## E=0 ## inside the conductor, so there is no contribution to the flux from the inside surface. Alternatively, if there is air on both sides, both sides get the same outward pointing ## E ## and the result is ## E=\frac{\sigma}{2 \epsilon_o} ##. ## \\ ## (In full detail, for the conductor case: ## \int E \cdot dA=EA=\frac{\sigma A}{\epsilon_o} ##, so that ## E=\frac{\sigma}{\epsilon_o} ##. ## \\ ## And for the case with air on both sides: ## \int E \cdot dA=2 E A=\frac{\sigma A}{\epsilon_o } ##, so that ## E=\frac{\sigma}{2 \epsilon_o} ## ).

- #6

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So, the solution to part A) is:A layer of surface charge where you have air on both sides when you draw what is called a Gaussian pillbox around a section of area ## A ## of surface charge density ## \sigma ##. When you calculate the flux on the "inside", the conductor necessarily has ## E=0 ## inside the conductor, so there is no contribution to the flux from the inside surface. Alternatively, if there is air on both sides, both sides get the same outward pointing ## E ## and the result is ## E=\frac{\sigma}{2 \epsilon_o} ##. ## \\ ## (In full detail, for the conductor case: ## \int E \cdot dA=EA=\frac{\sigma A}{\epsilon_o} ##, so that ## E=\frac{\sigma}{\epsilon_o} ##. ## \\ ## And for the case with air on both sides: ## \int E \cdot dA=2 E A=\frac{\sigma A}{\epsilon_o } ##, so that ## E=\frac{\sigma}{2 \epsilon_o} ## ).

and part C) is:

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It says it is a metallic sphere so isn't it a conductor?

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Yes, that means it's a conductor, but they did not tell you in part A to derive the Gauss' law for the case of conductor material on one side of the Gaussian pillbox. It is a very good result that they are having you derive, but they need to present the question with sufficient detail that you get the correct result. There are two different cases that can occur. The case with air on both sides is not relevant here, but if you just read part A, it would appear that's what they are asking for.It says it is a metallic sphere so isn't it a conductor?

- #10

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I see. Thank you!Yes, that means it's a conductor, but they did not tell you in part A to derive the Gauss' law for the case of conductor material on one side of the Gaussian pillbox. It is a very good result that they are having you derive, but they need to present the question with sufficient detail that you get the correct result. There are two different cases that can occur. The case with air on both sides is not relevant here, but if you just read part A, it would appear that's what they are asking for.

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