Using Gauss' law to find the induced surface charge density ##\sigma##

[Meow12]

Homework Statement

A thin insulating rod with charge density $\lambda=\rm+5\ nC/m$ is arranged inside a thin conducting cylindrical shell of radius $R=\rm 3\ cm$. The rod and shell are on the same axis, and you can assume they are both infinite in length. What is the SURFACE charge density $\sigma$ induced on the OUTSIDE of the conducting shell in $\rm C/m^2$?

Relevant Equations

Statement of Gauss's Law: $\displaystyle\oint\limits\vec{E}\cdot d\vec{A} = \frac{Q}{\epsilon_0}$

My attempt:

The electric field in the interior of a conductor is $0$.

By symmetry, the electric field is directed radially outward.

Take the Gaussian surface as the thin cylindrical shell of radius $\rm 3\ cm$ and length $L$.

$\displaystyle\oint\limits\vec{E}\cdot d\vec{A} = \frac{Q}{\epsilon_0}$

Since $E=0$ everywhere, $Q=0$

$\lambda L+\sigma\cdot 2\pi R L=0$

$\lambda+2\pi R\sigma=0$

$\displaystyle\sigma=\rm-\frac{\lambda}{2\pi R}$

Upon substituting the values, we get $\rm\sigma=-2.6\times 10^{-7}\ C/m^2$

$\sigma_{outside}=\rm+2.6\times 10^{-7}\ C/m^2=\rm +260\ nC/m^2$

But the correct answer is $\rm +26\ nC/m^2$. I'm off by a factor of $10$; where have I gone wrong?

 

[TSny]

Welcome to PF!

$\displaystyle\sigma=\rm-\frac{\lambda}{2\pi R}$

Upon substituting the values, we get $\rm\sigma=-2.6\times 10^{-7}\ C/m^2$

Your formula is correct. When I substitute the values, I get a result that is about 1/10 of your value. Check your work. If you still aren't getting the correct value, show the numerical values that you used in the formula.

 

[Meow12]

Welcome to PF!Your formula is correct. When I substitute the values, I get a result that is about 1/10 of your value. Check your work. If you still aren't getting the correct value, show the numerical values that you used in the formula.

Yeah, I had made a silly calculation mistake. Thanks for your post.