Relating Volume Charge Density and Linear Charge Density

  • #1

Homework Statement


An infinite line of charge with linear density λ1 = 6.7 μC/m is positioned along the axis of a thick insulating shell of inner radius a = 2.4 cm and outer radius b = 4.7 cm. The insulating shell is uniformly charged with a volume density of ρ = -722 μC/m3.

What is λ2, the linear charge density of the insulating shell?

Homework Equations


(Vouter shell-Vinner shell)*ρ <-- current one I'm using, but I could be wrong.

The Attempt at a Solution


My approach has been to find a bridge between the two densities by relating the volume to the length, but in every attempt I've made I've run into the problem of what to do about the length variable, as the shell is infinite. This is a gauss's law problem, so I've also tried using that to get rid of length variable - however I realized I didn't have the E-field.

With the current equation I'm using, I've again run into not knowing what to do about the length of the cylindrical shell.

ETA - Okay, I've now tried this:

ρ=Q/(2pi(b2-a2)L) and Q=ρ*V=λ*L

So, λ*L=(2pi(b2-a2)L)*ρ, and the lengths will cancel

λ=(2pi(b2-a2))*ρ

Answer is wrong though.
 
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Answers and Replies

  • #2
haruspex
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λ=(2pi(b2-a2))*ρ
Looks right to me. What do you get numerically? Do you know what the answer is supposed to be?
 
  • #3
lightgrav
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Homework Statement


An infinite line of charge with linear density λ1 = 6.7 μC/m is positioned along the axis of a thick insulating shell of inner radius a = 2.4 cm and outer radius b = 4.7 cm. The insulating shell is uniformly charged with a volume density of ρ = -722 μC/m3.

What is λ2, the linear charge density of the insulating shell?

Homework Equations


(Vouter shell-Vinner shell)*ρ <-- current one I'm using, but I could be wrong.

The Attempt at a Solution


My approach has been to find a bridge between the two densities by relating the volume to the length, but in every attempt I've made I've run into the problem of what to do about the length variable, as the shell is infinite. This is a gauss's law problem, so I've also tried using that to get rid of length variable - however I realized I didn't have the E-field.

With the current equation I'm using, I've again run into not knowing what to do about the length of the cylindrical shell.

ETA - Okay, I've now tried this:

ρ=Q/(2pi(b2-a2)L) and Q=ρ*V=λ*L

So, λ*L=(2pi(b2-a2)L)*ρ, and the lengths will cancel

λ=(2pi(b2-a2))*ρ

Answer is wrong though.
There is no "2" in the Volume of a cylinder ... it is Area x Length.
 
  • #4
haruspex
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There is no "2" in the Volume of a cylinder ... it is Area x Length.
Ah yes - I didn't notice the 2.
 
  • #5
Looks right to me. What do you get numerically? Do you know what the answer is supposed to be?
I did (-722E-6)(2pi(0.0472-0.0242)) = -7.41 μC, I input -7.41 as the units were already included and the answer was wrong. I have no access to the answer.
 
  • #6
Ah!!! I have no idea how I didn't notice that. Removed the 2 and the answer was correct. Thank you!

I must've carried the 2 around after playing with using the surface area to solve.
 

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