A duct's cross-section occupies the strip with earthed walls at and

PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Conductors, capacitance and boundary-value electrostatics · Conductors, capacitance and boundary-value electrostatics

The Boundary-Value Problem, Ch Separation of variables in Cartesian geometry

Problem

A duct's cross-section occupies the strip 0≤x≤a0\le x\le a with earthed walls at x=0x=0 and x=ax=a, and extends upward without limit in yy: there is no wall at any finite yy. The wall at y=0y=0 is held at some potential V(x,0)=f(x)V(x,0)=f(x) whose detailed shape does not matter here. Because the duct is open at large yy, the physical requirement replacing a fourth wall is that V(x,y)→0V(x,y)\to0 as y→∞y\to\infty for every xx in (0,a)(0,a), with VV bounded throughout. ((a)) Substitute V(x,y)=X(x)Y(y)V(x,y)=X(x)Y(y) and separate Laplace's equation, writing the two ordinary equations in terms of one constant μ\mu. ((b)) Using only the conditions X(0)=X(a)=0X(0)=X(a)=0 on the earthed side walls, show by testing both signs of μ\mu that μ\mu must be negative, and find the allowed values of kk where μ=−k2\mu=-k^2. ((c)) With μ=−k2\mu=-k^2, the equation for YY is Y′′=k2YY''=k^2Y, whose general solution is a sum of two exponentials. Show that only one combination is consistent with V→0V\to0 as y→∞y\to\infty, and write it down.

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