Calculating Energy of Coaxial Cylinders

In summary: I am just a little lost on how to solve this equation.Thanks for your help!In summary, Using gauss's law, the electric field between the two cylinders is given by \frac{λ}{2*pi*ε0*s} where s is the radius of the gaussian surface.
  • #1
thatguy14
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Homework Statement



There are 2 long coaxial insulating cylinder. The inner and outer cylinders have radii of a and b and charge densities λ and -λ uniformly distributed on the surface. Calulcate energy per unit length 2 ways (equations below)

Homework Equations



W = [itex]\frac{1}{2}[/itex][itex]\int σ V da[/itex]
W = [itex]\frac{ε0}{2}[/itex][itex]\int E^2 dτ[/itex]


The Attempt at a Solution



So using gausses law, with s (radius of cyliunder) < a there is no electric field as there is no enclosed charge. Outside both cylinders there is no E because there is no net charge. SO the only E and energy is between the 2 cylinders.

So I have to do it 2 ways, one with the first and one with the second. I am unsure though how to set up gauss's law to find the E field. Also since we are using line charge how does that change things?

So i wrote |E|* 2*pi*l*s =1/ε0[itex]\int λ*2*pi*s*dl[/itex]

but this doesn't seem right. I am unsure how to deal with the line charge. Do i use like surface charge still but write it as σ = 2*pi*r*dl but what are the limits of integration? I am just a little confused at this part. Help would be nice! Thanks
 
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  • #2
sorry I wiorked a little bit, is the electric field E = λ/(2*ε0) * (s^2-a^2) where s is the radius on between a and b?

Edit:

I did some looking around and worked a little more and here is where I am at. The electric field in between the two cylinders is [itex]\frac{λ}{2*pi*ε0 * s}[/itex] where s is the radius of the gaussian surface. So then the potential in this area is then

v = -[itex]\int E ds[/itex] and the limits of integration go from b to a (reference at infinity).
so then plugging in E we get

[itex]\frac{λ}{2*pi*ε0}[/itex]* ln(b/a) (negative was used to switch division).

Then do I plug this into the first formula for work? Does σda = λ*l*2*pi*s*ds? What are the limits of integration? From a to b? I think I made a mistake somwehre.
 
Last edited:
  • #3
Hi guys, so I figured out my main problem with this question. I am having troubles coming to terms with the fact that were using a cylinder but the charge is given in terms of the linear charge density. So when using gauss's law, what is the charge enclosed? Is it just lamba * l?
 

1. How do you calculate the energy of coaxial cylinders?

The energy of coaxial cylinders can be calculated using the equation: E = 0.5 * ε * A * (V1 - V2)^2, where ε is the permittivity of the material between the cylinders, A is the area of the cylinders, and V1 and V2 are the potentials of the inner and outer cylinders, respectively.

2. What is the formula for the area of coaxial cylinders?

The formula for the area of coaxial cylinders is A = 2πh * (r1 + r2), where h is the height of the cylinders and r1 and r2 are the radii of the inner and outer cylinders, respectively.

3. What does the permittivity represent in the energy calculation for coaxial cylinders?

The permittivity represents the ability of a material to store electric charge and is measured in Farads per meter (F/m). It is a constant that depends on the material between the cylinders.

4. Can the energy of coaxial cylinders be negative?

Yes, the energy of coaxial cylinders can be negative if the potential difference (V1 - V2) is negative, indicating that the inner cylinder has a higher potential than the outer cylinder. This can occur if the cylinders are connected to a power source in reverse polarity.

5. What factors can affect the energy calculation for coaxial cylinders?

The energy calculation for coaxial cylinders can be affected by the permittivity of the material between the cylinders, the potential difference between the cylinders, and the area of the cylinders. Additionally, the presence of other nearby conductors can also affect the energy calculation.

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