A question on E Flux and Fields

In summary, the field between two infinite charged sheets of opposite charge is σ/ε0 due to superposition. However, the same cannot be applied to a conducting sphere embedded in another with charges Q and -Q on the outer and inner surfaces respectively. This is because the field inside a spherically symmetric charge shell is zero. While attempting to solve this problem, the field of the inner cylinder can be found using flux, but there is no need to consider superposition with the field of the inner surface of the outer cylinder. The resulting field is just due to the inner cylinder with a value of E = Q/(2πrhε0).
  • #1
WWCY
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Homework Statement


1. The field for an infinite charged sheet is found to be σ/2ε0. If we place 2 infinite sheets of opposite charge above one another, we say that the field in between the sheets is σ/ε0 due to the superposition of individual fields.

Why can't we say the same for a situation where a conducting sphere is embedded in another, and there is charge Q on the outer surface on the inner sphere and -Q on the inner surface of the outer sphere? By same I mean applying superposition on fields by Q and -Q.

Homework Equations

The Attempt at a Solution


The only reason i could think of was that it was impossible (to me) to find a Gaussian surface capable of enclosing the charge on the inner surface on the outer sphere, which was hardly a rigorous explanation.

Any help is much appreciated!
 
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  • #2
The fields in the case of the spheres are also a superposition of the individual fields. It is just that the field inside a spherically symmetric charge shell is zero.
 
  • #3
Orodruin said:
The fields in the case of the spheres are also a superposition of the individual fields. It is just that the field inside a spherically symmetric charge shell is zero.

Apologies, but I might not have put the question across properly, it was in relation to this problem (part a).

I tried to solve by finding the field of the inner cylinder using flux, after which I was confused on whether or not i should be doing any "superposition" with the field by the inner surface of the outer cylinder. I then found out that that would've been a mistake, and the field was actually just due to the inner one.

E = Q/(2πrhε0)

Thank you for helping!
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1. What is E Flux and Fields?

E Flux and Fields is a scientific concept that describes the flow of energy and matter through a given area, as well as the associated electromagnetic fields. It is commonly used in the study of physics, astronomy, and other related fields.

2. How does E Flux and Fields affect our daily lives?

E Flux and Fields play a crucial role in our daily lives, as they are responsible for the functioning of electronic devices, communication systems, and power grids. They also play a role in natural phenomena such as lightning and the Earth's magnetic field.

3. What factors affect E Flux and Fields?

E Flux and Fields are affected by various factors, including the strength of the source of energy or matter, the distance from the source, and the medium through which the energy or matter is traveling. Additionally, the presence of other fields can also influence E Flux and Fields.

4. How is E Flux and Fields measured and calculated?

E Flux and Fields are typically measured using instruments such as Gaussmeters and electrometers that can detect and quantify the strength of magnetic and electric fields, respectively. These values can then be used in mathematical equations to calculate the total E Flux and Fields.

5. What are some real-world applications of E Flux and Fields?

E Flux and Fields have numerous practical applications, including in the fields of telecommunications, power generation and distribution, medical imaging, and space exploration. Understanding and manipulating E Flux and Fields is also crucial in developing sustainable energy sources and protecting against harmful radiation.

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