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!

I Induced surface charge density

  1. Feb 18, 2017 #1
    Hello! I am a bit confused about calculating the induced surface charge density on an infinite conducting plane, with 0 potential, in the presence of a charge, q, a distance d above it. Assuming that the plane is in xy plane and the charge in positive z region, in the book they use the method of mirror charges to calculate the potential V at every point in space and in order to calculate the surface charge density they use: ##\sigma=-\epsilon_0\frac{\partial V}{\partial z}_{(z=0)}##. How do they get to this formula?
    Thank you!
     
  2. jcsd
  3. Feb 18, 2017 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Take Gauss's Law (in SI units :nb))
    $$\vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho.$$
    On the surface you have a jump in the electric field since for ##z<0## it's identically 0 and for ##z>0## it's described by the potential
    $$V(\vec{r})=\frac{q}{4 \pi \epsilon_0 \sqrt{x^2+y^2+(z-d)^2}} -\frac{q}{4 \pi \epsilon_0 \sqrt{x^2+y^2+(z+d)^2}}, \quad z>0,$$
    as you should have figured out from the method of image charges.

    Now integrate Gauss's Law over a little cube parallel to the ##xy## plane around ##\vec{r}=(x,y,0)## on the surface. The total charge included is
    $$Q_V=\epsilon_0 E_z \Delta A,$$
    where ##\Delta A## is the area of the surface parallel to the ##xy## plane, where ##E_z## is to be taken at ##(x,y,0^+)##, because all other contributions of the surface integral cancel or the field is ##0## for ##z<0##. Thus at the end you get
    $$\sigma(x,y)=\epsilon_0 E_z(x,y,0^+)=-\epsilon_0 \partial_z V(x,y,0^+).$$
     
  4. Feb 18, 2017 #3
    Awesome! Thank you!
     
  5. Feb 19, 2017 #4
    Sorry, I have a question actually. Why is the electric field 0, for z<0?
     
  6. Feb 20, 2017 #5

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    That's obviously a solution of the boundary-value problem, and it's known to be unique. So this must be the solution for this electrostatic problem.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted