- #1
hectoryx
- 15
- 0
Homework Statement
The capacitor is assumed to consist of two circular disc electrodes of radius [tex]\alpha [/tex]. The electrodes are of infinitesimal thickness, placed a distance [tex]2L[/tex]
apart, and are equally and oppositely charged to potentials +V and -V. To solve the potential distribution in and out the capacitor, the Laplace Equation is:
[tex]\[\frac{{{\partial ^2}\phi }}{{\partial {r^2}}} + \frac{1}{r}\frac{{\partial \phi }}{{\partial r}} + \frac{{{\partial ^2}\phi }}{{\partial {z^2}}} = 0{\rm{ }}...{\rm{equ}}{\rm{.(1)}}\][/tex]
A solution in a paper to (1) is
[tex]\[\phi (r,z) = \int_0^\infty {A(\lambda )({e^{ - \lambda |z - L|}} - {e^{ - \lambda |z + L|}}){J_0}(\lambda r)} d\lambda {\rm{ }}...{\rm{equ}}{\rm{.(2)}}\][/tex]
I do not really understand why we can derive (2) from (1) directly. And my solution is a little different with (2). Does anyone tell me what is wrong with my method?
I solve (1) using separation of variables,
Homework Equations
Assume [tex]\[\phi (r,z) = R(r)Z(z)\][/tex] ,and divide through by [tex]\[\phi (r,z)\][/tex] ,
[tex]\[\frac{{Z(z)}}{r}\frac{{\partial R(r)}}{{\partial r}} + Z(z)\frac{{{\partial ^2}R(r)}}{{\partial {r^2}}} + R(r)\frac{{{\partial ^2}Z(z)}}{{\partial {z^2}}} = 0\][/tex]
So, we can get:
[tex]\[\frac{1}{Z}\frac{{{\partial ^2}Z}}{{\partial {z^2}}} = {\lambda ^2}{\rm{ }}...{\rm{equ}}{\rm{.(3)}}\][/tex]
[tex]\[\frac{1}{{Rr}}\frac{{\partial R}}{{\partial r}} + \frac{{{\partial ^2}R}}{{R\partial {r^2}}} = - {\lambda ^2}{\rm{ }}...{\rm{equ}}{\rm{.(4)}}\][/tex]
The Attempt at a Solution
The solution for (3) and (4) is
[tex]\[Z(z) = C(1){e^{z\lambda }} + C(2){e^{ - z\lambda }}\][/tex]
[tex]\[R(r) = C(3){J_0}(r\lambda ) + C(4){Y_0}{\rm{(}}r\lambda )\][/tex]
Using boundary conditions, we have
[tex]\[Z(z) = C(1)({e^{z\lambda }} - {e^{ - z\lambda }}){\rm{ }}...{\rm{equ}}{\rm{.(5)}}\][/tex]
[tex]\[R(r) = C(3){J_0}(r\lambda ){\rm{ }}...{\rm{equ}}{\rm{.(6)}}\][/tex]
Now we get
[tex]\[\phi (r,z) = R(r)Z(z) = \int_0^\infty {A(\lambda )({e^{z\lambda }} - {e^{ - z\lambda }}{\rm{)}}{J_0}(r\lambda )d\lambda } {\rm{ }}...{\rm{equ}}{\rm{.(7)}}\][/tex]
Equation (7) is my solution and is really different with (2). I do not know what is wrong with my method.
Besides, I slightly do not understand the [tex]$\int_0^\infty {} $[/tex], in (7), why not[tex]\[\sum\limits_{\lambda = - \infty }^\infty {} \]
[/tex] or [tex]\[\sum\limits_{\lambda = 0}^\infty {} \][/tex]?
Please give me some advice. Thanks in advance
Regards
Hector