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2-D Laplace equation in rectangular coordinates

  • #1

Homework Statement


"The potential in the x-z plane is independent of z and given by a repeating step-function of magnitude 2(phi_0) and period 2a. The plane at y = y_0 is held at ground potential. Find the potential in teh region 0 < y < y_0." - Marion and Heald Classical Electromagnetic Radiation 3rd ed.




Homework Equations


d2(X)/dx2*X^-1 + d2(Y)/dy2*Y^-1 = 0

The Attempt at a Solution


X(x) must be oscillatory, and I can determine X(x) by standard Fourier analysis of square wave.

If X(x) is oscillatory, d2(X)/dx2 = -k1* X -> d2(Y)/dy2 = +k1*Y, to meet laplacian(phi) = 0.

Thus, Y(y) is a linear combination of growing and decaying exponential functions in y. But, this can never satisfy Y(y_0)=0. This is my problem, am I missing something important?

Thanks,
 

Answers and Replies

  • #2
I was thinking more about the possiblity of meeting boundary condition y(y_0)=0 with linear combination of growing and decaying exponentials in y, and even if you make the coefficients in front of growing exponentials 0, and set D*E^-K1*y = 0 at y-y_0, this implies that K1-> Infinity, which yields trivial solution for phi, namely phi==0 (double equals means identically, don't know how else to put symbols). It seems the only way to satisfy the boundary is if k1 is imaginary, making oscillatory solutions in Y(y), so that E^k1* y can be negative, in general. Does this mean that the potential can NOT be independant of z, since then 1/Z*d2Z/dz2 = H^2, so that H=(k1^2+k2^2)^0.5, for 1/X*d2X/dx2 = -k2z??

Thanks
 
  • #3
Solved the problem!

Of course, the coefficient of the decaying exponential has to have opposite sign as that of growing exponential (sinh), and the argument is altered so that the argument is 0 at y=y_0, so you get 1 + -1 = 0 for Y(y).
 

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