Electric Field of a Cylinder (Hollow/Solid) at the face

[adenine7]

Homework Statement

Part a.
A cylinder (with no face) is centered symmetrically around the z axis, going from the origin to infinity. It has charge density σ and radius R. Find the electric field at the origin.

Part b.
Same problem, except this time instead of a hollow cylinder with no face we have a solid cylinder. It also now has charge density ρ. Find the electric field at the origin.

Homework Equations

E_ring = ^kQz/_{(z²+ R²)^3/2}

σ =^dq/_2πRdz

E_disk = ^σ/_2εo * (1 - ^z/_{(z²+R²)^1/2})

ρ =^dq/_πR²dz

3. The Attempt at a Solution
I understand that to complete part a I must integrate the electric field of a charged ring (a distance z from its center) from z = 0 to z = ∞. I understand that to complete part b I must integrate the electric field of a charged disk (a distance z from its center) from z = 0 to z = ∞. Here is my solution to part a (I'm not sure if it's correct):

E_{Hollow Cylinder} = ∫₀^{∞ kQz}/_{(z²+ R²)^3/2} dz = ^kσ2π/_R

I don't know how to set up the integral for part b. When I integrate the equation for E_disk, with a dz tacked onto the end (not so sure about this), I get undefined. Is this a part of the problem, or is my math incorrect?

 

[adenine7]

Anyone have any ideas?

 

[TSny]

E_{Hollow Cylinder} = ∫₀^{∞ kQz}/_{(z²+ R²)^3/2} dz = ^kσ2π/_R

If you check the dimensions of your answer, you can see that it has the wrong dimensions for electric field.

The integral should represent a "sum" of fields of infinitesimally thin rings. The charge in one of the rings will be infinitesimal and should be represented by dq rather than Q. You can express dq in terms of the the charge density σ and dz. (This is how dz gets into the integral.)

 

[adenine7]

Thanks for the reply!
I think I made a mistake typing that in; the equation I integrated (which I did not correctly transcribe, sorry for the confusion) is with dQ = σ2πRdz (that's why my final answer has σ2π in it).

Assuming this is the correct answer, here's where I get stuck with part b. Rather than being able to manipulate Q into an expression of σ and area, the equation for the electric field of a disk already has σ (which I assume can just be replaced with ρ). I don't see anything I can manipulate but the ρ to get a dz into the equation, but when I do that I get a dq and a dz.

 

[TSny]

Thanks for the reply!
I think I made a mistake typing that in; the equation I integrated (which I did not correctly transcribe, sorry for the confusion) is with dQ = σ2πRdz (that's why my final answer has σ2π in it).

Yes, dQ = σ2πRdz. But, your answer to part (a) is not correct. You can see that your result has the wrong dimensions for an electric field. I think you are probably just making a minor error in not keeping track of the factors of R.

Assuming this is the correct answer, here's where I get stuck with part b. Rather than being able to manipulate Q into an expression of σ and area, the equation for the electric field of a disk already has σ (which I assume can just be replaced with ρ). I don't see anything I can manipulate but the ρ to get a dz into the equation, but when I do that I get a dq and a dz.

You cannot replace σ with ρ. They do not have the same meaning or the same dimensions. If dQ represents the total charge in one of the disks of thickness dz, try expressing dQ in terms of ρ, R, and dz. Then, get an expression for σ in terms of ρ and dz by noting that σ = dQ/(area of disk).

 

[adenine7]

Part a.
If I integrate kσ2πRz/(z2+ R2)3/2 dz from 0 to ∞ I get kσ2π, which checks out with dimensions. I believe I was forgetting to integrate with the extra z in the numerator (which does not get replaced by dz).

Part b.
Thank you for explaining the difference between ρ and σ! I thought they were interchangeable.

dQ = ρπR²dz

σ = dQ / 2πR = ρπR²dz / 2πR = ρRdz / 2

However, integrating with this substitution from 0 to ∞ still gets me an undefined solution.

∫₀^∞ (ρR / 4ε_o) * (1 - z / (z²+R²)^1/2) dz = [(ρzr / 4ε_o) - (ρr (z² + r²)^1/2 / 4ε_o)]₀^∞

 

[TSny]

Part a.
If I integrate kσ2πRz/(z2+ R2)3/2 dz from 0 to ∞ I get kσ2π, which checks out with dimensions. I believe I was forgetting to integrate with the extra z in the numerator (which does not get replaced by dz).

OK. That looks good. Of course, you could express your answer in terms of ε₀ instead of k.

Part b.
Thank you for explaining the difference between ρ and σ! I thought they were interchangeable.

dQ = ρπR²dz

σ = dQ / 2πR = ρπR²dz / 2πR = ρRdz / 2

Is 2πR the correct expression for the area of the disk?

However, integrating with this substitution from 0 to ∞ still gets me an undefined solution.

∫₀^∞ (ρR / 4ε_o) * (1 - z / (z²+R²)^1/2) dz = [(ρzr / 4ε_o) - (ρr (z² + r²)^1/2 / 4ε_o)]₀^∞

To evaluate "at infinity" you need to take the limit of the expression as z → ∞. You should find that the limit is finite.

 

[adenine7]

Expressing my electric field for a hollow cylinder in terms of ε_o, I get σ / 2ε_o. So, this is the same as the electric field for an infinite sheet of charge at distance z away?

For part b., I changed the expression for σ (fixed the area of a disk) and got σ = ρdz. I did what you said and expressed the integral evaluation as E = lim_{b → ∞^-} [E(b) - E(0)]. I get E = ρR / 2ε_o. This checks for the dimensions of E, I believe (though clearly I'm prone to simple mistakes).

 

[TSny]

Expressing my electric field for a hollow cylinder in terms of ε_o, I get σ / 2ε_o. So, this is the same as the electric field for an infinite sheet of charge at distance z away?

Yes. That's an interesting coincidence!

For part b., I changed the expression for σ (fixed the area of a disk) and got σ = ρdz. I did what you said and expressed the integral evaluation as E = lim_{b → ∞^-} [E(b) - E(0)]. I get E = ρR / 2ε_o. This checks for the dimensions of E, I believe (though clearly I'm prone to simple mistakes).

That all looks correct to me. Good work.

 

[adenine7]

Thank you so much for your help! You're a great teacher.