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Image charge

  1. Nov 20, 2009 #1
    What is image charge? and how we can use image charge?
    thank you so much.
  2. jcsd
  3. Nov 20, 2009 #2


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    When a charge or distribution of charge is placed in proximity to a conductor, the electric field from the charge induces a charge distribution on the surface of the conductor. Normally, it may be a difficult mathematical exercise to find the charge distribution that is induced. This is desirable because we could then use the induced charge distribution in conjunction with the source charges to find the net electric field of the system.

    However, we can note that the electric field must satisfy certain boundary conditions at the surface of the conductor. If we can setup an effective source inside the conductor that satisfies the boundary conditions on the surface, then we can use the effective source to be the source for our electric fields, avoiding the messy math involved in finding the exact charge distribution and associated fields. The boundary conditions require that the tangential electric field is zero along the surface of the conductor. Thus, all we need is to create an "image" charge that replicates this. The idea behind this is called the equivalence principle.

    An easy example is with an infinite conducting plate. If we place a charge at a distance d above the plate, we need to have the tangential electric field zero on the plate's surface. This is easily replicated by placing an equal but opposite charge at a distance d below the plate's surface and below the original charge. This way, the distance from the original charge and the image charge is the same at any point on the plate's surface and since they are opposite charges only a normal net field exists. Thus, the total electric field is the electric field of the original and image charges. This is much easier to calculate than solving the equations to find the true induced charges on the surface of the conductor and then integrating to find the electric fields.

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  4. Nov 20, 2009 #3
    Please give me a problem that use the method of image charge.
  5. Nov 20, 2009 #4


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    I already did, the total field of a point charge above an inifinite conducting plate.
  6. Nov 20, 2009 #5


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    In addition to what Born2bwire said it should be mentioned that for a point charge q located in front of an infinite plate conductor the boundary conditions usually said to be:
    [tex]\varphi(x,y,0)=0[/tex] (1)
    (z is perpendicular to the plate), not the "tangential electric field is zero". The last one is actually a consequence of (1) (which is, of course, very obvious due to the symmetry of the problem and the fact that electric field is always perpendicular to the conductor). See question 4 for details.

    IPhO' 2008, to play with this problem you can try to find the answers for:

    1) Why [tex]\varphi(x,y,0)=0[/tex] for all [tex](x,y)[/tex] points on an infinite plate conductor?

    2) Choose an arbitrary point (x,y,z) in space (as in the first question x and y axis lies on the plate, z axis is perpendicular to the plate and connects the real charge [tex]q[/tex] and the image-charge [tex]q'[/tex], [tex]2L[/tex] is a distance between charges). Find the expression for the potential (according to what you find in the first question for [tex]\varphi[/tex] when [tex]x,y \to \infty[/tex]) in the arbitrary point (x,y,z).

    3) Finding the derivatives
    [tex]\frac{\partial \varphi(x,y,z)}{\partial x}, \frac{\partial \varphi(x,y,z)}{\partial y}, \frac{\partial \varphi(x,y,z)}{\partial z}[/tex]
    show that only the last one is not a zero for such (x,y,z) that z=0 (i.e. for all points on the plate).

    4) Recall that
    [tex]\mathbf{E}(x,y,z) = -grad~\varphi(x, y, z) = -\left( \boldsymbol{i}\frac{\partial \varphi}{\partial x} + \boldsymbol{j}\frac{\partial \varphi}{\partial y} + \boldsymbol{k}\frac{\partial \varphi}{\partial z} \right)[/tex]
    where [tex]\mathbf{i}, \mathbf{j}, \mathbf{k}[/tex] are the unit vectors for x, y and z axis respectively. Using this show that
    [tex]E_x = E_y = 0[/tex] and
    [tex]E_z = - \frac{qL}{2 \pi \varepsilon_0 \left( x^2 + z^2 + L^2 \right)^{3/2}}[/tex] (2)

    5) Let [tex]\theta[/tex] be the angle (the smallest one) between z-axis and the line from q charge to an arbitrary point (x,y) on the plate, so
    [tex]x^2 + y^2 = L^2 tan^2 \theta[/tex] (3)
    Show that (2) becomes
    [tex]E_z= \frac{q Cos^3 \theta}{2 \pi \varepsilon_0 L^2}[/tex]
    Using Gauss theorem prove that charge density on the plate as a function of theta is
    [tex]\sigma(\theta)=-\varepsilon_0 E_z = -\frac{q Cos^3{\theta}}{2 \pi L^2}[/tex](4)
    For this consider a wafer thin pillbox one half of which is inside the conductor and another half is outside of it.

    6) Now consider a little band on the plate between [tex]\theta[/tex] and [tex]\theta + d \theta[/tex]. Electric charge induced by q on this band is
    [tex]dQ = \sigma(\theta) dS[/tex] = [tex]\pi \sigma(\theta) d(r^2)[/tex]
    where r is the radius of the band. Since [tex]x^2 + y^2 = r^2[/tex] from (3) one can derive:
    [tex]dQ = \pi L^2 \sigma(\theta) d(tan^2 \theta)[/tex]
    Substituting (4) into this equation and integrating over all theta show that total induced charge on the surface is -q:
    [tex]Q = -q[/tex]

    7) q charge is attracted toward the plane. Find the force between q and conductor using Coulomb law.

    8) Show that the energy (potential energy) of two point charges q and -q without conductor (i.e. in vacuum) is
    [tex]W= -\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{2 L}[/tex] (5)
    (calculate the work required to bring q from infinity to (0,0,L) when -q is fixed at (0,0,-L)).

    9) Explain why the energy of a single charge q and conducting plate is only half of (5)?

    I think it's a good way to start with these little problems before solving much more complicated. When you'll finish with these I can give you another three or four.
    Last edited: Nov 20, 2009
  7. Nov 20, 2009 #6
    If you want a good selection of 2D image charge problems, look at Smythe "Static and Dynamic Electricity" 3rd Edition Chapter IV. For example, consider a charge Q at an arbitrary point between two mutually perpendicular conducting sheets.
    Bob S
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