Electric Field due to a charged conducting finite cylindrical shell.

In summary, the conversation discusses the difficulties in finding the electric field for a cylindrical capacitor using Gauss's law and integrating the field between two conducting cylindrical shells. The question of whether Gauss's law can be applied to a finite cylinder is raised, and it is noted that the field lines are radial even at the end of the cylinder. The formula for the field is given, but there is confusion about the potential being infinity at infinity. It is then clarified that Gauss's law cannot be used for a finite charged cylinder, and the conversation concludes with a discussion about calculating the field and using it to find the capacitance.
  • #1
ed2288
25
0
Hi everyone. I'm having a bit of trouble with finding an electric field. Basically, I'm trying to understand the formula for a cylindrical capacitor, so the method involves integrating the field between two conducting cylindrical shells. Firstly can Gauss's law be used in this case, because the cylinder is finite? Are the field lines all radial, even at the very end of the cylinder? If so, this leads to my next problem. The field turns out to be:
(charge/2*Pi*Length*epsilon_0*radius)
So, when you integrate this to obtain the potential, you will end up with a natural logarithm, meaning at infinity, the potential is infinity!? I'm sure this is wrong but I just can't see where the error is. Any help would be greatly appreciated!
 
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  • #2
The Gauss theorem CANNOT be used for finite charged cylinder.
 
  • #3
Ok then, but how can you calculate the field inbetween the two finite cylinders? You need the field so you can integrate it to get the potential, which you can then use to calculate the capacitance, which, I'm told, turns out to be
C=(2*Pi*epsilon_0*Length)/log((second radius)/(first radius))
 

1. What is an electric field?

An electric field is a physical quantity that describes the strength and direction of the force that a charged particle experiences when placed in the vicinity of other charged particles. It is represented by vector arrows, with longer arrows indicating a stronger field.

2. How is the electric field calculated due to a charged conducting finite cylindrical shell?

The electric field due to a charged conducting finite cylindrical shell can be calculated using the formula E = (σ/2ε0)(1 - (R1/R)), where σ is the surface charge density of the cylinder, ε0 is the permittivity of free space, R1 is the radius of the cylinder, and R is the distance from the center of the cylinder to the point where the electric field is being calculated.

3. What is the relationship between the electric field and the surface charge density of a cylindrical shell?

The electric field is directly proportional to the surface charge density of a cylindrical shell. This means that if the surface charge density increases, the electric field will also increase, and vice versa. This relationship is represented by the formula E ∝ σ.

4. How does the electric field due to a charged conducting finite cylindrical shell vary with distance from the cylinder?

The electric field due to a charged conducting finite cylindrical shell follows an inverse relationship with distance. This means that as the distance from the cylinder increases, the electric field decreases. This is represented by the formula E ∝ 1/R, where R is the distance from the center of the cylinder to the point where the electric field is being calculated.

5. Can the electric field due to a charged conducting finite cylindrical shell be zero?

Yes, it is possible for the electric field due to a charged conducting finite cylindrical shell to be zero. This occurs when the distance from the center of the cylinder to the point where the electric field is being calculated is equal to the radius of the cylinder, or when R = R1. In this case, the electric field is canceled out due to the opposite charges on the inner and outer surfaces of the cylinder.

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