1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding surface charge densities and potential (metal cylinder)

  1. Oct 13, 2014 #1
    1. The problem statement, all variables and given/known data

    A solid metal cylinder of radius ##R## and length ##L##, carrying a charge ##Q##, is surrounded by a thick coaxial metal shell of inner radius ##a## and outer radius ##b##. The shell carries no net charge.
    a) Find the surface charge desnities ##\sigma## at ##R##, at ##a##, and at ##b##.
    b) Find the potential at the center using ##r=b## as the reference point.
    2. Relevant equations
    ##\sigma=q/A##, ##A_{cyl}=2\pi rh##, ##V(r)=-\int_{r_1}^{r_2} \frac{\sigma}{r} da##

    3. The attempt at a solution
    a) ##\sigma_R=\frac{Q}{2\pi RL}## ##\sigma_a=\frac{Q}{2\pi aL}## ##\sigma_b=\frac{Q}{2\pi bL}##

    b) ##V=-\int_b^a \frac{\sigma_b}{r}da-\int_a^R \frac{\sigma_a}{r}da-\int_R^0 \frac{\sigma_R}{r}da##
    ##da=2\pi RLdr##
     
  2. jcsd
  3. Oct 14, 2014 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The last equation is not correct. How are the potential function V(r) and the electric field E(r) related ?

    You should integrate with respect to the variable the integrand depends on. If you integrate a function depending on r , write it as ##\int f(r)dr ## instead of ##\int f(r) da##. If you use fixed boundaries in the integral for r the result can not depend on r .
    Moreover, a means the inner radius of the shell, do not use it as an integrating variable.

    The surface charge densities are all right, but the sign of the surface charge density at a is not correct.
    I do not understand what you tried to say with line b). Does the potential change inside a conductor?

    ehild
     
  4. Oct 14, 2014 #3
    I should've made it made it more clear that ##da## is the normal vector of the surface area of the cylinder so that ##d\vec{a}=2 \pi RLdr\hat{r}##

    Yeah I forgot the minus sign. Also the potential does not change in the conductor so I should remove the integral that goes from 0 to ##R##
     
    Last edited: Oct 14, 2014
  5. Oct 15, 2014 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Note that the potential is line integral of the negative electric field instead of a surface integral.

    That is right, but it is zero also somewhere else.

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Finding surface charge densities and potential (metal cylinder)
Loading...