- #1
davidbenari
- 466
- 18
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##.
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##.
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