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!

Homework Help: Charge above and infinite grounded plane

  1. Jun 20, 2012 #1
    The classic image problem is this: Suppose we have a charge q over an infinite grounded plane. What will be the potential at a given point for z>0. It's not just the potential of the charge q since it will induce a certain charge on the conductor.
    By looking at the mirror problem of a charges of opposite sign and displaced the same length relative to z=0 but in different directions, one finds from the uniqueness theorem that this solution also holds for the problem above.
    Now: According to my book, even though the fields will the same, then the electrostatic energy of the charge is different in the two problems.
    In the problem with the conductor the energy is namely only half the energy of the image problem. I have some trouble understanding this. The electrostatic energy is negative because we have to do work to separate the system in both cases. In the first case a negative charge and positive charge are brought in. In the second case a charge q is brought in while the induced charge runs in to sustain a zero potential in the conductor. But why would separation require half the energy?
    In the first case you would need to pull away the positive charge away while keeping the negative fixed. In the conductor case, what would then happen?
  2. jcsd
  3. Jun 21, 2012 #2


    User Avatar
    Homework Helper
    Gold Member

    This easiest way to see this is to look at the expression for electrostatic energy in term of the electric field [tex]W=\frac{\epsilon_0}{2}\int_{\text{all space}} E^2 d^3 x[/tex]

    The fields for [itex]z\geq0[/itex] are the same for both cases, but what about for [itex]z<0[/itex]?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook