# Relating Volume Charge Density and Linear Charge Density

## 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

Last edited:

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

lightgrav
Homework Helper

## 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

There is no "2" in the Volume of a cylinder ... it is Area x Length.

haruspex
Homework Helper
Gold Member
There is no "2" in the Volume of a cylinder ... it is Area x Length.
Ah yes - I didn't notice the 2.

Looks right to me. What do you get numerically? Do you know what the answer is supposed to be?