Potential and E Field due to a point charge near a grounded conducting right angle

PROBLEM:

A point charge Q is located in the vicinity of  a grounded conducting "right angle" with infinite planar electrodes in the x >0, y > 0 half spaces as shown.  Q lies in the x-y plane at distances (Subscript[d, x]) and (Subscript[d, y]) from the two planes respectively.  We wish to determine the electric potential and field in the region where x >0 and y > 0.  Realize  the conductors shown in the figure extend to infinity in the ± z directions (and the +x, +y directions).

(a)  What classic technique or method that is used for determining V and E for charges near certain symmetric geometries of conductors is your likely choice??

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(b)  In this type of problem, what role do boundary conditions play in finding V and E?

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It is understood that the field of a point charge (∝(1/r)) satisfies LE.
We can quickly use M to verify by using the Laplacian Function in Spherical Coordinates.  If we get 0, we’re done — using a superposition of point charges (one real; others image charges), we will have a solution to Laplace’s Equation:

(*Execute code below*)
Laplacian[(1/r),{r,θ,ϕ},"Spherical"]  (* V of a single point charge ∝ 1/r *)

So, Your Response included  Images charges??   Brilliant!   The challenge is to find a set of point charges that will result in the electrodes along the x and y axes to remain at zero volts (ground).  Hints:  

1.  We use the results of using a single image charge for the case of a single real charge Q a distance +dy in front of an infinite grounded conducting (x-z) plane (the correct image charge, Qimage= –Q, placed a distance (–) dy behind the plane).  This will yield a zero potential along the entire x-z plane.

Single Plane; point charge:
Reminder:  To find V[x,y,z] for y > 0 ONLY, we take simply add V(Q) + V(-Q);  

					i.e.,    V[x,y,z] = k Q( (1/(Sqrt[(x^2) + (((y -d))^2) + (z^2)]))  + (1/(Sqrt[(x^2) + (((y + d))^2) + (z^2)]))  )

2.  A combination of 3 image charges with magnitudes ± Q (and properly positioned) will yield infinite (x-z) and (y-z) planes at zero potential .  Here is what the planes look like AND where (approximately) the charges might be placed.

(c)  Determine the magnitudes and positions of the image charges such that these planes are all at zero volts potential.  [Superhint: The image charges are all in the same plane as Q ( the z = 0 plane).]

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(d)  Derive the Potential V[x,y,z]  (i.e., V at the points P {x,y,z} ) in the region x>0, y>0).  (Easiest to use M to get the results.)

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ClearAll["`*"] (*Leave ClearAll["`*"] statement*)

(e)  Use M to show that the BCs of the problem are satisfied.

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V[0,y,z]
V[x,0,z]

Voila!  We see that the original BCs are satisfied.  Thus, our V is THE solution.

(f) Plot V[x,y,0] for both + and – (x and y), realizing that the actual region of interest corresponds to x > 0, y > 0 [this contains the point charge (lying in the z = 0 plane)].  (You want to use Plot3D.)

We have to give the parameters actual numerical values;  I use max = 4 as a coordinate max in both x and y for all of the plots.  That is, both x and y range from –max to +max.

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(g)  Interpret your plot.

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(h)  Now do a Plot of V in the (x>0, y>0) sector at various z (z moves you above or below the z = 0 plane which contains Q), using Manipulate

Bigus Hintus:   Here’s the code:  

Manipulate[Plot3D[V[x,y,z],{x,0,max},{y,0,max},PlotRange-> {0,5} ],{z,-1,1}]   

(The “{0,5}” simply scales the V axis conveniently.)
NOTE:  when you manipulate z, M may take a few seconds to re-plot V.

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(i)  Interpret your results.

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If we set x = (Subscript[d, x]), y = (Subscript[d, y]) (exactly above or below Q) and then plot V vs. z we see how the maximum value of V (for a given z) falls with |z|.

(*Excecute code below*)
Plot[V[dx,dy,z],{z,-5,5}]

Since this is the MAX of V for a given z, we see that V rapidly diminishes at all points in this segment.    At z = 0 you are at the position of the real point charge Q and therefore V = ∞.

(j) Now find E (suggest you call it EE) in this segment. (Hint:  use –Grad[] in Cartesian Coordinates).

(*Input code below*)
Clear[k , Q , dx, dy , max ] (*Leave Clear[] command.*)
V[x,y,z];
EE[x_,y_,z_] = - Grad[V[x,y,z],{x,y,z}]//Simplify

It’s not pretty but we see that it has three components {EEx, EEy, EEz}.  

For a plot we will do in the near-future, I want to show that in the z = 0 plane, EEz = 0;  we set z = 0 and pick out the zth (3rd) component of EE:

(*Execute code below*)
EE[x,y,0][[3]]

(k)  Now do a VectorPlot3D (limited to our "real sector", x > 0, y > 0);  we can let z go from some –zmax to +zmax for illustration.

Because Q is contained in the region of the plot, the infinity in EE at it’s location {+(Subscript[d, x]), +(Subscript[d, y]), 0} causes M great PAIN in any VectorPlot.  Setting EE = {0,0,0} at points near the position of Q helps M to behave; I also use VectorScale to play with the relative lengths of the vectors.  

Use your mouse to rotate the plot to view these vectors from different vantage points.

(*Execute code below*)
k = 1; Q = 1; dx=1; dy = 2; max = 4;
EEE[x_,y_,z_] = If[Norm[EE[x,y,z]] > 0.4,{0,0,0},EE[x,y,z]];
VectorPlot3D[EEE[x,y,z],{x,0, max},{y,0, max},{z,-max,max},VectorScale->0.09,AxesLabel->{"x", "y", "z"} ]

(l) Interpret your VectorPlot3D.

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We can get a different perspective by using StreamPlot in 2D.  Above we showed EEz = 0 by setting z = 0 and picking out the zth (3rd) component of EE:

EE[x,y,0][[3]] = 0.

Here is the StreamPlot of EE in the z = 0 plane:

(*Execute code below*)
StreamPlot[{EE[x,y,0][[1]],EE[x,y,0][[2]]},{x,0,max},{y,0, max},FrameLabel->{"x", "y"} ]

Remember the Streamlines show the flow of EE; magnitudes are not indicated. We made Q positive, so EE flows out of Q.  We see clearly that EE is normal to both the x = 0 and y = 0 planes (grounded electrodes).  Not at all obvious, BUT:  all field lines leaving Q will terminate at one of the two electrodes.  

Now look at the induced charge distributions on the two electrode surfaces.  You recall that at a conducting surface, σ = ((Subscript[E, ⊥]^surface)/(ϵo)).  
It’s not hard to see that (Subscript[E, ⊥]^surface) at the x = 0 plane is EE[0,y,z][[1]]  and (Subscript[E, ⊥]^surface) at the y = 0 plane is EE[x,0,z][[2]].  So we can quickly generate the charge densities on the two half plane electrodes in the problem ( I call them  sigmax = “sigma_x = 0_plane” and sigmay = “sigma_y = 0_plane”):

(m)  Now plot the induced charge distributions on the two electrode surfaces.  Recall that at a conducting surface,

σ = ((Subscript[E, ⊥]^surface)/(Subscript[ϵ, 0]))   .

It’s not hard to see that (Subscript[E, ⊥]^surface) at the x = 0 plane is EE[0,y,z][[1]]  and (Subscript[E, ⊥]^surface) at the y = 0 plane is EE[x,0,z][[2]].  So we can quickly generate the charge densities on the two half plane electrodes in the problem (I call them  sigmax = “sigma_x = 0_plane” and sigmay = “sigma_y = 0_plane”).

(*Execute code below*)
Clear[k , Q , dx, dy , max ] (*Leave Clear[] command*)
sigmax= ((EE[0,y,z][[1]])/(ϵo))
sigmay  = ((EE[x,0,z][[2]])/(ϵo))
k = 1; Q = 1; dx=1; dy = 2;max = 4; ϵo =1;
sigmax
Plot3D[sigmax,{y,0,max},{z,-max,max},PlotRange->{0,-2},AxesLabel->{"y", "z","sigma_x = 0_plane"}]
Plot3D[sigmay,{x,0,max},{z,-max,max},PlotRange->{0,-1},AxesLabel->{"x", "z","sigma_y = 0_plane"}]

(n)  Interpret the surface charge density plots.

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ONE MORE TIME:  V and E in the x>0, y > 0 sector is due to the V and E from Q plus these two charge distributions BUT, as mentioned already, we can get V and E quick like bunnies by using the image charges determined above.  

It’s a circular argument in the sense that we used the image charges to calculate V; this gave us E via –Grad[V], which allowed us to calculate the two sigmas.  

The bottom line is that for the calculations:  IMAGE CHARGES.   For understanding the resulting electrostatics:  the +Q induces negative surface charges (electrons pulled up onto the electrodes from the ground); the resulting V and E are due to the combination of the +Q and the negative sigmas.  Note that there are no other “real” charges involved.  The metal electrodes cannot contain charge inside the metal.  The negative σ’s reside on the surfaces towards the +Q.