- #1
hectoryx
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Homework Statement
The capacitor is assumed to consist of two parallel circular disc electrodes of radius R. The electrodes are of infinite small thickness, placed a distance 2H apart, and are equally and oppositely charged to potentials +U and -U. A metal cylinder is placed near the two electrodes and the position relationship can be found in the following picture:
http://i1021.photobucket.com/albums/af335/hectoryx/professional/1.jpg
Homework Equations
To solve the potential distribution in this situation, the Laplace Equation in cylindrical coordinate system is:
[tex]\[{\nabla ^2}\phi = \frac{1}{r}\frac{{\partial \phi }}{{\partial r}} + \frac{{{\partial ^2}\phi }}{{\partial {r^2}}} + \frac{{{\partial ^2}\phi }}{{\partial {z^2}}} = 0\][/tex]
The Attempt at a Solution
I am not sure about its boundary condition:
[tex]\[\left\{ {\begin{array}{*{20}{c}}
{\phi = + {\rm{U}},\begin{array}{*{20}{c}}
{} & {{\rm{z}} = {\rm{H}}} \\
\end{array},0 \le r \le {\rm{R }}} \\
{\phi = - {\rm{U}},\begin{array}{*{20}{c}}
{} & {{\rm{z}} = - {\rm{H}}} \\
\end{array},0 \le r \le {\rm{R}}} \\
\end{array}} \right.\][/tex]
and
[tex]\[\begin{array}{l}
\phi = {\phi _c},\begin{array}{*{20}{c}}
{} & {{\rm{H}} + {\rm{d}} \le {\rm{z}} \le {\rm{H}}} \\
\end{array} + {\rm{d}} + {\rm{L}},0 \le r \le {\rm{R}} \\
\frac{{\partial \phi }}{{\partial z}} = {\sigma _1},\frac{{\partial \phi }}{{\partial r}} = 0,\begin{array}{*{20}{c}}
{} & {{\rm{z}} = {\rm{H}} + {\rm{d}}} \\
\end{array},0 \le r \le {\rm{R}} \\
\frac{{\partial \phi }}{{\partial z}} = {\sigma _2},\frac{{\partial \phi }}{{\partial r}} = 0,\begin{array}{*{20}{c}}
{} & {{\rm{z}} = {\rm{H}} + {\rm{d}}} \\
\end{array} + {\rm{L}},0 \le r \le {\rm{R}} \\
\frac{{\partial \phi }}{{\partial z}} = 0,\frac{{\partial \phi }}{{\partial r}} = {\sigma _3},\begin{array}{*{20}{c}}
{} & {{\rm{H}} + {\rm{d}} \le {\rm{z}} \le {\rm{H}}} \\
\end{array} + {\rm{d}} + {\rm{L}},r = {\rm{R}} \\
\end{array}\]
[/tex]
Could anyone give me some help and tell me that whether the boundary conditions above are right?
Thanks very much!
Best Regards.
Hector