Laplace Equation on semi-infinite plate

AI Thread Summary
A problem involving a thin semi-infinite conductive plate with a voltage applied at one end was discussed, where the goal was to find the voltage V(x) at a distance x from the end. The initial solution, V(x)=-90*x+1, did not satisfy the boundary conditions at infinity, while the correct solution, V(x)=V0*10^(-x/d), did. The discussion highlighted that electrostatic potential satisfies Laplace's equation in regions with zero charge density and Poisson's equation otherwise. The realization that the problem involves a conductor clarified why the correct solution does not satisfy Laplace's equation. The charge density as a function of x was also mentioned as epsilon_0*(ln10)^2*exp(-x/d).
ZombieCat
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Hello all!

I just finished the following problem:

Consider a thin semi-infinite plate of negligible thickness made of an isotropic conductive material. A voltage V0=1V is applied at x=0 on the plate (across the short dimension). At a distance x=d=1cm from the end (x=0) V is measured to be .1V. Find the voltage V(x) at an arbitrary distance x from the end.

In my first attempt I got V(x)=-90*x+1, which is a solution to the Laplace equation in 1D, but does not match the boundary condition at infinity.

I tried the problem again and got V(x)=V0*10^(-x/d), which matches all boundary conditions and is the correct answer. My question is why doesn't this solution satisfy the Laplace equation? Does it have to? Why/why not?
 
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The electrostatic potential only satisfies Laplace's equation in regions where the charge density is zero. The rest of the time it satisfies Poisson's equation.
 
Oh dangit! Cause we're dealing with a conductor gotcha! So I guess the epsilon_0*(ln10)^2*exp(-x/d) would be the charge density as a function of x.

Thanks!

*Maybe I should rename myself "the phorgetful physicist"...
 
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