- #1
Telemachus
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Hi there. The example to obtain the electric potential in the region z>0 by the method of images for a infinite grounded conducting plane, with a point charge q located at a distance d is a typical example of the application of the method of images.
If we consider that the plane is located in the xy plane, at z=0, and the charge at z=d, we must fulfill the boundary conditions:
1. ##V=0## when ##z=0##
2. ##V \rightarrow 0## far from the charge, that is for ##x^2+y^2+z^2>>d^2##.
So locating the image charge at z=-d one can get the desired potential for the region of space of interest, being:
##\displaystyle V(x,y,z)=\frac{1}{4\pi \epsilon_0}\left [ \frac{q}{\sqrt{x^2+y^2+(z-d)^2}}-\frac{q}{\sqrt{x^2+y^2+(z+d)^2}} \right ]##
Now, if the plane is not grounded, how should I modify this to set it, let's say at a potential ##V_0##?
I thought that one posibility would be to use an other image distribution, an infinite charged plane parallel to the conducting plane, at a position ##-z_0## such that ##\frac {\sigma}{2\epsilon_0}z_0=V_0##, where ##\sigma## is the charge per unit area in the charged plane. I think that another possibility would be just to put the charges over the conducting plane. Now the induced charges will be, ofcourse infinite. And the electric field will be the sum of the images plus the real charge in the region z>0.
So, what I should have would be something like:
##\displaystyle V'(x,y,z)=\frac{1}{4\pi \epsilon_0}\left [ \frac{q}{\sqrt{x^2+y^2+(z-d)^2}}-\frac{q}{\sqrt{x^2+y^2+(z+d)^2}} \right ]+\frac{V_0}{z_0}(z_0-z)##
The potential due to the infinite charged plane should grow with z, and at the same time it should meet the condition of V(x,y,z=0)=V_0. I'm having trouble with where I'm puting the origin of the potential with respect to the two planes I'm considering.
Thanks in advance.
If we consider that the plane is located in the xy plane, at z=0, and the charge at z=d, we must fulfill the boundary conditions:
1. ##V=0## when ##z=0##
2. ##V \rightarrow 0## far from the charge, that is for ##x^2+y^2+z^2>>d^2##.
So locating the image charge at z=-d one can get the desired potential for the region of space of interest, being:
##\displaystyle V(x,y,z)=\frac{1}{4\pi \epsilon_0}\left [ \frac{q}{\sqrt{x^2+y^2+(z-d)^2}}-\frac{q}{\sqrt{x^2+y^2+(z+d)^2}} \right ]##
Now, if the plane is not grounded, how should I modify this to set it, let's say at a potential ##V_0##?
I thought that one posibility would be to use an other image distribution, an infinite charged plane parallel to the conducting plane, at a position ##-z_0## such that ##\frac {\sigma}{2\epsilon_0}z_0=V_0##, where ##\sigma## is the charge per unit area in the charged plane. I think that another possibility would be just to put the charges over the conducting plane. Now the induced charges will be, ofcourse infinite. And the electric field will be the sum of the images plus the real charge in the region z>0.
So, what I should have would be something like:
##\displaystyle V'(x,y,z)=\frac{1}{4\pi \epsilon_0}\left [ \frac{q}{\sqrt{x^2+y^2+(z-d)^2}}-\frac{q}{\sqrt{x^2+y^2+(z+d)^2}} \right ]+\frac{V_0}{z_0}(z_0-z)##
The potential due to the infinite charged plane should grow with z, and at the same time it should meet the condition of V(x,y,z=0)=V_0. I'm having trouble with where I'm puting the origin of the potential with respect to the two planes I'm considering.
Thanks in advance.
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