0
$\begingroup$

I'm currently studying electromagnetism, specifically Gauss's Law, and have been presented with the following question:

Consider an infinitely long cylinder of radius R made out of a conducting material. The charge density of the surface of the cylinder is 𝜎. Use Gauss law to calculate the electric field outside the cylinder. (Note that the element of surface in cylindrical coordinates is given by π‘‘π‘Ž = π‘ π‘‘πœ™π‘‘π‘§).

I am still quite stuck despite having searched the internet for a walkthrough of this problem. The answers I can find do not seem to contain the surface charge Οƒ term which leads me to believe my answer is wrong. My current working out is as followed:

Gauss's Law is: $$ \oint E da = \frac{Q_{enc}}{\epsilon_0} \label{eq1} $$

Using a cylindrical Gaussian surface with radius r coaxial with the infinitely long conducting cylinder of radius R and length l, I calculated the LHS of the above equation to be: $$ \oint E da = |E|2\pi r l \label{eq2} $$ Then working out the RHS as I understand it gives me: $$ \frac{Q_{enc}}{\epsilon_0} = \frac{Οƒa}{\epsilon_0} = \frac{Οƒ2\pi R l}{\epsilon_0} $$ Equating the LHS to the RHS and rearranging for $E$: $$ |E|2\pi r l = \frac{Οƒ2\pi R l}{\epsilon_0} $$ $$ E = \frac{R\sigma}{r\epsilon_0} $$ Which just does not sit right with me. Where have I gone wrong? I feel like I am being a bit thick at the moment. The website Hypherphysics (http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecyl.html) states that the electric field outside the conducting infinite cylinder should be: $$ E = \frac{\lambda}{2\pi r\epsilon_0} $$ However it doesn't explain how they arrived at that answer, whilst also not containing $\sigma$. Any help would be greatly appreciated! Hopefully I have made this clear enough a question.

$\endgroup$
  • $\begingroup$ As $\lambda = 2\pi R \sigma$ the two answers are identical. $\endgroup$ – my2cts Apr 9 at 12:48
2
$\begingroup$

Your solution is actually correct. The difference between your solution and the one you quote is that $\lambda/(2\pi \epsilon r)$ is the field of a wire, not that of a cylinder of finite radius $R$. For a wire one usually gives a linear charge density $\lambda$ but for a conducting cylinder this charge is on its surface, so the charge is given through a surface charge density $\sigma$. The total charge enclosed by your Gaussian cylinder (of radius $r$ and length $\ell$) is the charge on the surface of your conducting cylinder (of length $\ell$ and radius $R<r$): $$ Q=\sigma \times 2\pi R\times \ell\, . $$ Note that, for the wire version of this problem, $Q=\lambda\times \ell$ and you would recover the expression given in your link.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

perhaps your answer is wrong ,please check correctly

the electric field outside any conductor surface (in your question charges are at surface) is $\cfrac{\sigma}{\epsilon o}$

in non uniform charge distribution $\sigma$ may be different at different points but question states uniform surface charge distribution

link describes cylinder of volume charge density not surface one like in your question

this is true for every case in electrostatics, the electric field outside surface of conductor is as above

as electric field inside conductor is $0$ , your question states a cylinder made of conducting material

in a conductor charges always accumulate at surface to make flux $0$ inside since $E$$=$ $0$ (inside conductor) so flux should be $0$ as well.

so the link you have mentioned doesn't mention conducting word for cylinder of surface charge density as in your question

question states surface charge density $\sigma$ not volume charge density like in link

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.