# Electromagnetism and a solid metallic sphere

• Jozefina Gramatikova
In summary, the electric field on a surface of a conductor is dependent on whether there is a conductor on one side or free space on both sides. Gauss' law explains that the electric flux is equal to the enclosed charge divided by the permittivity of free space. When there is a conductor on one side, the electric field is equal to the surface charge density divided by the permittivity of free space. When there is free space on both sides, the electric field is equal to half the surface charge density divided by the permittivity of free space. Understanding Gauss' law is key to deriving these results.

## The Attempt at a Solution

The textbook says that the electric field on a surface of a conductor is:
. So, I guess since the sphere is metallic I can assume that what I have written there is true?

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The answer for the electric field ## E ## of a flat sheet with surface charge density ## \sigma ## needs some qualification here. A surface charge density of ## \sigma ## gives the result by Gauss' law that ## E=\frac{\sigma}{\epsilon_o} ## if there is a conductor on one side, but otherwise the result is ## E=\frac{\sigma}{2 \epsilon_o} ## if there is free space on both sides. It is the first case that applies here. ## \\ ## Gauss' law says that the electric flux which is ## \int E \cdot dA=\frac{Q_{enclosed}}{\epsilon_o} ##. It is difficult to explain the idea of flux without a chalkboard, but perhaps what I have given you will be helpful. ## \\ ## If you understand Gauss' law, deriving the result that ## E=\frac{\sigma}{\epsilon_o} ## at the surface of a conductor is very straightforward.

Jozefina Gramatikova
The answer for the electric field ## E ## of a flat sheet with surface charge density ## \sigma ## needs some qualification here. A surface charge density of ## \sigma ## gives the result by Gauss' law that ## E=\frac{\sigma}{\epsilon_o} ## if there is a conductor on one side, but otherwise the result is ## E=\frac{\sigma}{2 \epsilon_o} ## if there is free space on both sides. It is the first case that applies here. ## \\ ## Gauss' law says that the electric flux which is ## \int E \cdot dA=\frac{Q_{enclosed}}{\epsilon_o} ##. It is difficult to explain the idea of flux without a chalkboard, but perhaps what I have given you will be helpful. ## \\ ## If you understand Gauss' law, deriving the result that ## E=\frac{\sigma}{\epsilon_o} ## at the surface of a conductor is very straightforward.
what do you mean by "if there is free space on both sides"?

Jozefina Gramatikova said:
what do you mean by "if there is free space on both sides"?
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} ## ).

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Jozefina Gramatikova
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} ## ).
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} ## ).
So, the solution to part A) is:

and part C) is:
?

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Jozefina Gramatikova said:
So, the solution to part A) is: View attachment 229715
and part C) is: View attachment 229716 ?
Part A needs to be described in more detail, and they need to specify "with a conductor on one side". Without that additional detail being included, part C does not follow. I do think this my have been an oversight on the part of the person who made up the question.

Part A needs to be described in more detail, and they need to specify "with a conductor on one side". Without that additional detail being included, part C does not follow. I do think this my have been an oversight on the part of the person who made up the question.
It says it is a metallic sphere so isn't it a conductor?

Jozefina Gramatikova said:
It says it is a metallic sphere so isn't it a conductor?
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.

Jozefina Gramatikova
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.
I see. Thank you!

## 1. What is electromagnetism?

Electromagnetism is a branch of physics that deals with the interactions between electrically charged particles. It explains how electric and magnetic fields are generated and how they interact with each other.

## 2. How does a solid metallic sphere interact with electromagnetism?

A solid metallic sphere can interact with electromagnetism in various ways. It can act as a conductor, allowing electric charges to flow through it. It can also be magnetized by an external magnetic field, and it can generate its own magnetic field when an electric current flows through it.

## 3. What is the relationship between electricity and magnetism in a solid metallic sphere?

The relationship between electricity and magnetism in a solid metallic sphere is described by Maxwell's equations. These equations show that a changing electric field can create a magnetic field, and a changing magnetic field can create an electric field. This phenomenon is known as electromagnetic induction.

## 4. How is the strength of the magnetic field around a solid metallic sphere determined?

The strength of the magnetic field around a solid metallic sphere is determined by the amount of electric current flowing through it and the distance from the sphere. The closer the distance, the stronger the magnetic field will be. The strength of the magnetic field also depends on the material of the sphere, as some materials are better conductors than others.

## 5. Can a solid metallic sphere be both a conductor and an insulator?

No, a solid metallic sphere cannot be both a conductor and an insulator. A conductor allows electric charges to flow through it easily, while an insulator does not. Metallic spheres are typically good conductors, but they can become insulators if they are coated with an insulating material or if they are very small in size.