Gauss's Law problem involving a Cylinder

  • #1

Homework Statement


The figure a shows a narrow charged solid cylinder that is coaxial with a larger charged cylindrical shell. Both are non-conducting and thin and have uniform surface charge densities on their outer surfaces. Figure b gives the radial component ##E## of the electric field versus radial distance ##r## from the common axis, and ##E_{s}=3.0\times 10^{3}N/C##. What is the shell's linear charge density?

image.png


Homework Equations


##\oint \vec{E}\cdot\vec{dA}=\frac{q_{encl}}{\varepsilon_0}##

The Attempt at a Solution


First I derived the relationship between ##E## and ##\lambda##(linear charge density), i.e.
$$
\oint \vec{E}\cdot\vec{dA}=\frac{q_{encl}}{\varepsilon_0}\Longrightarrow E(2\pi rL)=\frac{\lambda L}{\varepsilon_0}\Longrightarrow E=\frac{\lambda}{2\pi r \varepsilon_0}
$$
Note:Using r=3.5cm to solve everything
From the left of r=3.5cm only the field from the first cylinder will be involved so we have $$E_1=\frac{\lambda_1}{2\pi r\varepsilon_0}=1000N/C$$
Now from the right of r=3.5 using the relation ##E_T=E_1+E_2##, ##E_2=E_T-E_1=-2000-1000=-3000N/C##, then going back to the first relation,
$$
E_T=\frac{\lambda_T}{2\pi r\varepsilon_0}=E_1+E_2\Longrightarrow \lambda_T=2\pi r\varepsilon_0(E_1+E_2)
$$
EDIT: Okay so I was trying to calculate the wrong thing, I should be trying to find the linear charge density of the larger shell, therefore I have ##E=\frac{\lambda}{2\pi r\varepsilon_0}##, first shell: ##E_1=\frac{\lambda_1}{2\pi r\varepsilon_0}##, both shells: ##E_T=\frac{\lambda_1}{2\pi r\varepsilon_0}+\frac{\lambda_2}{2\pi r \varepsilon_0}##, ##E_T-E_c=\frac{\lambda_2}{2\pi r\varepsilon_0}##, then I have $$(E_T-E_c)2\pi r\varepsilon_0=\lambda_2$$ which gives the negative of the answer so I'm not entirely sure what went wrong.
 
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Answers and Replies

  • #2
ehild
Homework Helper
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Homework Statement


The figure a shows a narrow charged solid cylinder that is coaxial with a larger charged cylindrical shell. Both are non-conducting and thin and have uniform surface charge densities on their outer surfaces. Figure b gives the radial component ##E## of the electric field versus radial distance ##r## from the common axis, and ##E_{s}=3.0\times 10^{3}N/C##. What is the shell's linear charge density?

image.png


Homework Equations


##\oint \vec{E}\cdot\vec{dA}=\frac{q_{encl}}{\varepsilon_0}##

The Attempt at a Solution


First I derived the relationship between ##E## and ##\lambda##(linear charge density), i.e.
$$
\oint \vec{E}\cdot\vec{dA}=\frac{q_{encl}}{\varepsilon_0}\Longrightarrow E(2\pi rL)=\frac{\lambda L}{\varepsilon_0}\Longrightarrow E=\frac{\lambda}{2\pi r \varepsilon_0}
$$
Note:Using r=3.5cm to solve everything
From the left of r=3.5cm only the field from the first cylinder will be involved so we have $$E_1=\frac{\lambda_1}{2\pi r\varepsilon_0}=1000N/C$$
Now from the right of r=3.5 using the relation ##E_T=E_1+E_2##, ##E_2=E_T-E_1=-2000-1000=-3000N/C##, then going back to the first relation,
$$
E_T=\frac{\lambda_T}{2\pi r\varepsilon_0}=E_1+E_2\Longrightarrow \lambda_T=2\pi r\varepsilon_0(E_1+E_2)
$$
EDIT: Okay so I was trying to calculate the wrong thing, I should be trying to find the linear charge density of the larger shell, therefore I have ##E=\frac{\lambda}{2\pi r\varepsilon_0}##, first shell: ##E_1=\frac{\lambda_1}{2\pi r\varepsilon_0}##, both shells: ##E_T=\frac{\lambda_1}{2\pi r\varepsilon_0}+\frac{\lambda_2}{2\pi r \varepsilon_0}##, ##E_T-E_c=\frac{\lambda_2}{2\pi r\varepsilon_0}##, then I have $$(E_T-E_c)2\pi r\varepsilon_0=\lambda_2$$ which gives the negative of the answer so I'm not entirely sure what went wrong.
Your derivation is correct, if Ec means E1. What result have you got?
 
  • #3
Your derivation is correct, if Ec means E1. What result have you got?
When I plug in numbers I obtain ##\lambda_2=5.83\times10^{-9}##, the solution in the book is given as ##\lambda_2=-5.83\times10^{-9}##, another odd thing is if I solve for the values of ##\lambda## then use ##\lambda_1+\lambda_2=\lambda_t## I get the correct answer but I would think that both methods should work.
 
  • #4
ehild
Homework Helper
15,543
1,914
When I plug in numbers I obtain ##\lambda_2=5.83\times10^{-9}##, the solution in the book is given as ##\lambda_2=-5.83\times10^{-9}##, another odd thing is if I solve for the values of ##\lambda## then use ##\lambda_1+\lambda_2=\lambda_t## I get the correct answer but I would think that both methods should work.
What is ET? Is not it negative?:smile:
 
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  • #5
What is ET? Is not it negative?:smile:
haha yea it appears as though I accidentally swapped the values for ##E_T## and ##E_1## when performing the calculation.
 

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