User:PAR/Work2

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[edit] Half plane coordinates

The potential near the φ=0 grounded half plane at [x1,y1,z1] with charge c at [xc1,yc1,zc1] is (Andrews 200???)

V=\frac{c}{\pi\sqrt{\rho_1\rho_{c1}}}\,\left[S(\chi,\phi_1-\phi_{c1})-S(\chi,\phi_1+\phi_{c1})\right]

where

\rho_1^2=x_1^2+y_1^2\,
\rho_{c1}^2=x_{c1}^2+y_{c1}^2\,
\varphi_1=\arctan^*(y_1,x_1)\,
\varphi_{c1}=\arctan^*(y_{c1},x_{c1})\,
\chi=\frac{\rho_1^2+\rho_{c1}^2+(z_1-z_{c1})^2}{2\rho_1\rho_{c1}}= \frac{r_1^2+r_{c1}^2-2z_1z_{c1}}{2\rho_1\rho_{c1}}

where the arctan* function returns a value between 0 and 2π

[edit] Image Sphere Coordinates

Covert to imaging sphere coordinates, subscript o. Imaging sphere has radius A

x_1=z_o\,
y_1=x_o-A^2/2R\,
z_1=y_o\,
x_{c1}=z_{co}\,
y_{c1}=x_{co}-A^2/2R\,
z_{c1}=y_{co}\,
r_1^2=r_o^2-\frac{A^2x_o}{R}+\frac{A^4}{4R^2}\,
r_{c1}^2=r_{co}^2-\frac{A^2x_{co}}{R}+\frac{A^4}{4R^2}\,
\rho_1^2 =z_o^2 +x_o^2 -\frac{A^2x_o}{R} +\frac{A^4}{4R^2}\,
\rho_{c1}^2=z_{co}^2+x_{co}^2-\frac{A^2x_{co}}{R}+\frac{A^4}{4R^2}\,
\varphi_1=\arctan^*(x_o-A^2/2R,z_o)\,
\varphi_{c1}=\arctan^*(x_{co}-A^2/2R,z_{co})\,
\chi_{num}=r_o^2-\frac{A^2x_o}{R}+r_{co}^2-\frac{A^2x_{co}}{R}-2y_oy_{co}+\frac{A^4}{2R^2}

[edit] Take the Image

Take the image by

r_o\rightarrow A^2/r_o

Thus:

x_o\rightarrow A^2x_o/r_o^2\,
y_o\rightarrow A^2y_o/r_o^2\,
z_o\rightarrow A^2z_o/r_o^2\,

The image potential is now

W=\frac{1}{r_o}V(r_o\rightarrow A^2/r_o)

and:

\rho_1^2\rightarrow\frac{A^4}{r_o^4}(z_o^2+x_o^2)-\frac{A^4x_o}{r_o^2R}+\frac{A^4}{4R^2}\,
\rho_{c1}^2\rightarrow z_{co}^2+x_{co}^2-\frac{A^2x_{co}}{R}+\frac{A^4}{4R^2}
\varphi_1\rightarrow\arctan^*(x_o-r_o^2/2R,z_o)\,
\varphi_{c1}\rightarrow\arctan^*(x_{co}-A^2/2R,z_{co})\,
\chi_{num}\rightarrow\frac{A^4}{r_o^2}-\frac{A^4x_o}{r_o^2R}+r_{co}^2-\frac{A^2x_{co}}{R} -\frac{2A^2y_oy_{co}}{r_o^2}+\frac{A^4}{2R^2}

[edit] Express charge position in terms of hemisphere charge position

Express in terms of the hemisphere charge location. Charge c is at \mathbf{r}_{co}=[x_{co},z_{co},z_{co}] so charge q is at \mathbf{r}_{qo}=(A^2/r_{co}^2)[x_{co},y_{co},z_{co}]=[x_{qo},y_{qo},z_{qo}]. Thus A2 = rcorqo and

x_{co}=\frac{A^2x_{qo}}{r_{qo}^2}
y_{co}=\frac{A^2y_{qo}}{r_{qo}^2}
z_{co}=\frac{A^2z_{qo}}{r_{qo}^2}

So that now:

\rho_1^2\rightarrow\frac{A^4}{r_o^4}(z_o^2+x_o^2)-\frac{A^4x_o}{r_o^2R}+\frac{A^4}{4R^2} =\frac{A^4}{r_o^4}\left(z_o^2+\left(x_o-\frac{r_o^2}{2R}\right)^2\right)\,
\rho_{c1}^2\rightarrow\frac{A^4}{r_{qo}^4}(z_{qo}^2+x_{qo}^2)-\frac{A^4x_{qo}}{Rr_{qo}^2}+\frac{A^4}{4R^2} =\frac{A^4}{r_{qo}^4}\left(z_{qo}^2+\left(x_{qo}-\frac{r_{qo}^2}{2R}\right)^2\right)\,
\varphi_1\rightarrow\arctan^*(x_o-r_o^2/2R,z_o)\,
\varphi_{c1}\rightarrow\arctan^*(x_{qo}-r_{qo}^2/2R,z_{qo})\,
\chi_{num}\rightarrow \frac{A^4}{r_o^4}\left(z_o^2+\left(x_o-\frac{r_o^2}{2R}\right)^2\right) +\frac{A^4}{r_{qo}^4}\left(z_{qo}^2+\left(x_{qo}-\frac{r_{qo}^2}{2R}\right)^2\right) -\frac{2A^4y_oy_{qo}}{r_o^2r_{qo}^2}

[edit] Convert to hemisphere coordinates

x_o=x+R\,
y_o=y\,
z_o=z\,
x_{qo}=x_q+R\,
y_{qo}=y_q\,
z_{qo}=z_q\,
r_o^2=r^2+2xR+R^2\,
r_{qo}^2=r_q^2+2x_qR+R^2\,
\rho_1^2=\frac{A^4}{r_o^4}\left(z^2+\left(\frac{R^2-r^2}{2R}\right)^2\right)=\frac{A^4}{r_o^4}\gamma_1^2
\rho_{c1}^2=\frac{A^4}{r_{qo}^4}\left(z_q^2+\left(\frac{R^2-r_q^2}{2R}\right)^2\right)=\frac{A^4}{r_{qo}^4}\gamma_{c1}^2
\chi=\frac{r_{qo}^4\gamma_1^2+r_q^4\gamma_{c1}^2-2yy_qr_o^2r_{qo}^2}{2\gamma_1\gamma_{c1}r_o^2r_{qo}^2}

Note that the A4 terms cancel in χ and the angles. The potential is

W=\frac{A}{r_o}\,\frac{c}{\pi\sqrt{\rho_1\rho_{c1}}}\,\left[S^--S^+\right]

using the charge for the hemisphere c = qA / rqo, it can be seen that the A's cancel here too, leaving a function that is independent of A.

W=\frac{q}{\pi\sqrt{\gamma_1\gamma_{c1}}}\,\left[S^--S^+\right]

PROBLEM - If this is correct, it is not in the best form. There should be an obvious symmetry in the x and y coordinates. Without loss of generality, we could have required the half-plane charge c to have zc1=0, so that in the hemisphere system yq=0 and the yyq cross term in the above expression for χ disappears. Now for a charge q off of the y=0 plane, just choose a coordinate system rotated about the z axis.

[edit] Charge along z axis

For a charge on the z axis

x_q=y_q=0\,
r_q=z_q\,
r_{qo}^2=R^2+z_q^2\,


and so:

\gamma_1^2=\left(z^2+\left(\frac{R^2-r^2}{2R}\right)^2\right)
\gamma_{c1}^2=\frac{r_{qo}^4}{4R^2}
\chi=\frac{r_{qo}^4\gamma_1^2+r_q^4\gamma_{c1}^2}{2\gamma_1\gamma_{c1}r_o^2r_{qo}^2} =\frac{4R^2\gamma_1^2+z_q^4}{4R\gamma_1r_o^2}=\frac{R\gamma_1}{r_o^2}+\frac{z_q^4}{4R\gamma_1r_o^2}