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Two questions related to "classical problem" of the image method

  1. Sep 19, 2015 #1
    The classical problem of the image method is:

    An infinite conducting and grounded (V=0) plate is on the xy plane. A charge ##q## is above it (we can think that it lies on the #z# axis). Knowing that ##V \to 0## as you move far away from the charge and that ##V=0## on the plane find the potential for the region ##z>0##.

    Okay and my questions are:

    (1) After finding ##V## why would ##\sigma = -\epsilon_o \frac{\partial V}{\partial n} \big|_{z=0}## be the surface charge on the plane?

    My objection to this is that the derivative ##\frac{\partial V}{\partial n}## is discontinuous at ##z=0##. It could either be ##0## or ##\sigma##. What makes it not vanish? In other words, why, if there is a discontinuity, does only the value we want pop up?

    (2) How can we be sure the electric field below the xy plane is ##0##?

    I think the typical answer is that fields can't penetrate conductors. Why is this the case? I do accept that ##E=0# inside a 3D conductor but this is a 2D conductor, and for that matter it seems to me that the statement "Fields can't penetrate conductors" is more general. Is it true? If not, then why is the field below the xy plane equal to zero?

    EDIT
    Mhmm, Three questions, sorry.

    (3) The energy can be calculated considering how much work it takes to bring the charge from infinity to a distance ##d## above the plate. Why is this so?

    My objection to this is that the calculation of energies for a continuous charge distribution and a collection of point charges if fundamentally different.

    The energy for a continuous charge distribution is positive definite since it is related to the integral of ##E^2## while the energy of a collection of particles can be negative. So it seems to me that we have to ways of talking about the energy here: Either we integrate the field over all space, or we just consider the work it takes to bring in charge ##q##.
     
    Last edited: Sep 19, 2015
  2. jcsd
  3. Sep 19, 2015 #2
    Let ##\Omega## be the region ##z>0##. We want to solve the following boundary value problem(s):
    $$\Delta \Phi_1 = - \rho \quad \text{ in } \Omega$$
    $$\Phi_1(z=0)=0$$
    where ##\rho=-q \delta(z-d)##
    And let ##\Omega^{'}## be the region ##z<0##
    $$\Delta \Phi_2 = 0 \quad \text{ in } \Omega^{'}$$
    $$\Phi_2(z=0)=0$$
    Where ## \vec E = -\nabla \Phi##

    For the solution in all ## \mathbb R^{3}## we must patch together the solutions for ##\Phi## in ##\Omega## and ##\Omega^{'}##
    In ##\Omega^{'}## the solution is clearly ##\Phi_2 \equiv 0 \Rightarrow \vec E_2 \equiv 0##, and we can understand that as if we have an infinite conductor sitting at ##z<0##

    Now, we can compute $$\sigma= \epsilon_0(\vec E_2 - \vec E_1) . \hat n \Rightarrow \sigma=- \epsilon_0 \frac{\partial \Phi}{\partial n}$$
     
    Last edited: Sep 19, 2015
  4. Sep 19, 2015 #3
    But ##\frac{\partial \Phi}{\partial n}## is discontinuous at the the plate isnt it? So it doesnt make much sense to evaluate it.
     
  5. Sep 19, 2015 #4
    No, the electric field is discontinuous at ##z=0##, but the potential is not.
    The solution (in cylindrical coordinates) is $$ \Phi_1(r,\phi, z) = \frac{1}{4 \pi \epsilon_0 } \bigg ( \frac{q}{\sqrt{r^2 + (z-d)^2}} - \frac{q}{\sqrt{r^2 + (z+d)^2}} \bigg ) \quad \quad z>0$$ $$\Phi_2 = 0 \quad \quad z<0$$
    Then $$\sigma = - \epsilon_0 \frac{\partial \Phi}{\partial z}(z=0) = \frac{-qd}{2 \pi (r^2 + d^2)^{3/2}}$$
     
    Last edited: Sep 19, 2015
  6. Sep 19, 2015 #5
    But if ##\vec{E_{above}}-\vec{E_{below}}=\sigma/\epsilon_o## then ##\frac{\partial \Phi}{\partial n}\Big|_{justabove} - \frac{\partial \Phi}{\partial n}\Big|_{justbelow} = -\sigma / \epsilon_o ## and the potential is discontinuous at z=0.

    What am I doing wrong then?
     
  7. Sep 19, 2015 #6
    The ##\underline{derivative}## of the potential in the normal direction (which is proportional to the normal component of the electric field) is discontinuous at ##z=0##

    You can check from the solution that i posted earlier that ##\Phi_2(z=0) = \Phi_1(z=0)## (i'm being inconsistent with the subscripts, but 2 stands for quantities in z<0 and 1 for quantities in z>0)
     
    Last edited: Sep 19, 2015
  8. Sep 19, 2015 #7
    Ohh you're right I meant the derivative all this time. :confused:

    So my question should really be why does ##\sigma = -\epsilon_o \frac{\partial \Phi}{\partial n}## if ##\frac{\partial \Phi}{\partial n} ## is discontinuous at z=0?

    I think thats the way I phrased it originally in my OP.
     
    Last edited: Sep 19, 2015
  9. Sep 19, 2015 #8
    The discontinuity of ##\vec E## (and hence the discontinuity of ##\frac{\partial \Phi}{\partial n}##) ##\underline{\text{at the boundary}}## is crucial in order to have a non-vanishing surfarce charge.
    Note that, if ##\frac{\partial \Phi}{\partial n}## were continuous, ##\sigma## would be identically zero.
    Note also that, although ##\frac{\partial \Phi}{\partial n}## is discontinuous at the boundary, ##\Phi## is not, thus the derivative is (piecewise) well defined.
    Also the derivatives ##\frac{\partial \Phi}{\partial n}## are evaluated ##\underline{\text{at the boundary}}##, so they have no dependency in z (##\sigma## is only defined at the boundary, the plate z=0): $$\sigma = - \epsilon_0 \frac{\partial \Phi}{\partial z}(z=0) = \frac{-qd}{2 \pi (r^2 + d^2)^{3/2}}$$
    I
     
    Last edited: Sep 19, 2015
  10. Sep 19, 2015 #9
    Without referring specifically to this problem, in theory, ##\frac{\partial \Phi}{\partial n}## will be discontinuous at the boundary. To picture another example: It would make no sense to evaluate ##\partial_n \Phi ## at the boundary of a spherical conductor, wouldn't it?

    I think what is making things work in this specific problem is that we are differentiating the potential which is valid for ##z>0##, and thus when we evaluate ##\partial_n \Phi## using this potential we are really doing ##\partial_n \Phi'\Big|_{justabove}## where ##\Phi'## is the general potential which describes the plate and the empty space below (note that the image isn't below the plate here) . And this would work because ##\vec{E}_{above}-\vec{E}_{below}= \sigma / \epsilon_o \hat{n} ##

    Could this be it?

    edit: I edited a lot of poorly made sentences.
     
  11. Sep 19, 2015 #10
    How can the derivative be well defined at that point if it is discontinuous at that point?
     
  12. Sep 19, 2015 #11
    (Yes it would work in the case of a spherical conductor. (Check Jackson's Classical Electrodynamics 3rd edition, p 58))
    Perhaps if I state all the equations it will become clearer. The solution of the problem is:
    $$ \Phi(r,\phi, z) \left\{
    \begin{array}{l}
    \Phi_1=\frac{1}{4 \pi \epsilon_0 } \bigg ( \frac{q}{\sqrt{r^2 + (z-d)^2}} - \frac{q}{\sqrt{r^2 + (z+d)^2}} \bigg ) \quad \quad z>0\\
    \ \Phi_2= 0 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \: \: z<0
    \end{array}
    \right.$$
    And the normal derivative:
    $$ - \nabla \Phi . \hat z = - \frac{\partial \Phi}{\partial z} =
    \left\{
    \begin{array}{l}
    \frac{\partial \Phi_1}{\partial z} = \frac{q}{4 \pi \epsilon_0} \bigg ( \frac{(z+d)}{(r^2+(z+d)^2)^{3/2}} - \frac{(z-d)}{(r^2+(z-d)^2)^{3/2}} \bigg ) \quad \quad z>0\\
    \ \frac{\partial \Phi_2}{\partial z}= 0 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \: \: z<0
    \end{array}
    \right.$$

    Now, if we want the surfarce charge:
    $$ \sigma= \epsilon_0(\vec E_2 - \vec E_1) . \hat n \quad \text{at z=0}= \epsilon_0 \bigg ( \lim_{z\rightarrow 0^{-}} {\frac{\partial \Phi_2}{\partial z}} - \lim_{z\rightarrow 0^{+}} {\frac{\partial \Phi_1}{\partial z}} \bigg ) \quad \\ = \frac{-qd}{2 \pi (r^2 + d^2)^{3/2}} $$
     
    Last edited: Sep 19, 2015
  13. Sep 19, 2015 #12
    I should have been clearer. When I said "well defined" i meant to say piecewise well defined, in the sense that the limit in the definition of the derivative exists, but is different depending on whether z<0 or z>0.
     
    Last edited: Sep 19, 2015
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