Gauss's Law problem involving a Cylinder

[Potatochip911]

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?

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.

 

[ehild]

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?

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?

 

[Potatochip911]

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.

 

[ehild]

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 E_T? Is not it negative?

 

[Potatochip911]

What is E_T? Is not it negative?

haha yea it appears as though I accidentally swapped the values for $E_T$ and $E_1$ when performing the calculation.