Concentric Cylindrical Conducting Shells and Capacitors

[diethaltao]

Homework Statement

An infinitely long solid conducting cylindrical shell of radius a = 3.1 cm and negligible thickness is positioned with its symmetry axis along the z-axis as shown. The shell is charged, having a linear charge density λinner = -0.49 μC/m. Concentric with the shell is another cylindrical conducting shell of inner radius b = 15.6 cm, and outer radius c = 18.6 cm. This conducting shell has a linear charge density λ outer = 0.49μC/m.

1.What is V(c) – V(a), the potential difference between the the two cylindrical shells?
2. What is C, the capacitance of a one meter length of this system of conductors?

Homework Equations

E = Q/(2πr^2εo)
Q = λSa where Sa = 2πr^2

The Attempt at a Solution

For the first question, I did the integral
∫E.dA = q/εo
E(2πr^2) = q/εo
E = Q/(2πr^2εo)

Then I integrated that between upper bound a and lower bound b to get:
ΔV = -Q/(2πεo) ∫dr/(r^2)
= Q/(2πεo)*(1/r) or Q/(2πεo)*((1/a)-(1/b))

However, my answer was incorrect. I am just learning integrals, so I'm not sure if I set up the integral incorrectly. The Q I used is the inner charge, calculated with Q = λ*(2πr^2), which, with values, is Q = (-0.49 μC/m)*(2π*0.031^2).
I've done a couple of problems with concentric insulators and conductors inside, but the whole idea is very confusing for me and I don't really understand concepts. I find myself just plugging numbers into random equations to find answers.

For the second question, would I just take the total Q and divide that by the potential difference?

Any help is appreciated. Thanks!

 

[BruceW]

Hi, welcome to physicsforums! Yeah, these integration problems can be tricky the first few times you do them.

E = Q/(2πr^2εo) This is meant to be the electric field due to a point charge, right? It should be 4pi, not 2pi. Also, you don't need this equation for this question. (You have started the attempt by using integration over a Gaussian surface, and I think this is the easiest way).

Q = λSa where Sa = 2πr^2 I don't think this is right. The question gives λ as the linear charge density, and the hint is in the name (and the units of λ). The cylinder extends indefinitely in the z-direction, and this is the reason that we require the use of λ in this problem. So what do you think you should multiply by λ to get Q?

∫E.dA = q/εo This is the right place to start. We want to find an equation for the electric field between the two cylinders, and considering the symmetry of the problem, what kind of Gaussian surface is the most convenient? (Hint: think about the direction of the electric field and what its magnitude depends on).

 

[diethaltao]

Thanks!

λ is charge per unit length or coulombs per meter, correct? I know that λL = Qenclosed, but I'm not sure what to do if it is infinitely long. Or does the L cancel off if we write:
E2πrL = Qenclosed/εo = λL/εo
Which can be simplified to:
E = λ/(2πrεo)?
And I could take the integral of λ/(2πr) to get:
λ/(2πεo)*ln(b/a)? Or am I completely off-track?

Wouldn't the Gaussian surface just be a cylinder? Or is it that because we treat it as a point charge, we use a sphere? The electric field is pointing radially inwards because the inner charge is negative, and magnitude depends on how far from the cylindrical shells the point is, right?

 

[BruceW]

λ is charge per unit length or coulombs per meter, correct? I know that λL = Qenclosed, but I'm not sure what to do if it is infinitely long. Or does the L cancel off if we write:
E2πrL = Qenclosed/εo = λL/εo
Which can be simplified to:
E = λ/(2πrεo)?
And I could take the integral of λ/(2πr) to get:
λ/(2πεo)*ln(b/a)? Or am I completely off-track?

This is completely correct. You seem to know this stuff pretty well :) But watch out about the sign, because λ is negative, and the gradient of the potential is negative of the electric field, so there is a chance to slip up on getting the sign correct.

Wouldn't the Gaussian surface just be a cylinder? Or is it that because we treat it as a point charge, we use a sphere? The electric field is pointing radially inwards because the inner charge is negative, and magnitude depends on how far from the cylindrical shells the point is, right?

Yep, the most convenient Gaussian surface is a cylinder (which is the surface you have used). You could also use a sphere, since you can actually use any kind of surface you like. But if you used a sphere, then the electric field on the surface would not be constant and wouldn't have constant direction, so the integral would be much more difficult. (And it would be more difficult to calculate the charge enclosed).

Also, yes, the electric field points inwards toward the z axis because the inner charges are negative, and the magnitude depends on how far the point is from the negatively charged inner cylindrical shell. These two facts are the reasons why using a cylindrical Gaussian surface is far more convenient.

Often when students first hear about using Gaussian surfaces in electromagnetism, they think something like "But there's no physical surface there, why should we integrate over a surface?" But mathematically, it is justified. And intuitively, I think of it like "how much electric field is flowing out of the closed volume", so in this sense the charges are a source of the electric field.

 

[diethaltao]

Thank you so much!
All of these integrals and concentric shells are really confusing.

If I just wanted to find the electric potential on the outer surface of the cylindrical shell, would I just do λ/(2πεo)*ln(a)? Or would I have to account for the charge of the charge of the surrounding conducting shell as well?

 

[BruceW]

hmm. You've already worked out that:
V_{(r_2)} - V_{(r_1)} = - \ \frac{\lambda}{2 \pi \epsilon_0} ln(\frac{r_2}{r_1})
Is the equation for the electric potential difference between two points which are in the space between the two metal cylinders. There is also an important rule for the electric potential within a conductor which is in equilibrium. (You probably know it, but you might have just forgotten it can be used in this situation). So from this, you should be able to work out the electric potential at any given point.

 

[diethaltao]

Ah, ok. I understand this much more. Hopefully I'll do well on my midterm tomorrow.

Thanks again!