Laplace Equation on semi-infinite plate

Click For Summary
SUMMARY

The discussion focuses on solving the Laplace equation for a thin semi-infinite plate made of isotropic conductive material with a voltage of V0=1V applied at x=0. The correct voltage distribution V(x) is derived as V(x)=V0*10^(-x/d), which satisfies the boundary conditions. The initial incorrect solution, V(x)=-90*x+1, fails to meet the boundary condition at infinity. The conversation highlights the distinction between Laplace's equation and Poisson's equation, emphasizing that electrostatic potential satisfies Laplace's equation only in charge-free regions.

PREREQUISITES
  • Understanding of Laplace's equation and Poisson's equation
  • Familiarity with electrostatics and electric potential
  • Knowledge of isotropic conductive materials
  • Basic calculus for solving differential equations
NEXT STEPS
  • Study the derivation and applications of Laplace's equation in electrostatics
  • Explore Poisson's equation and its implications in regions with charge density
  • Investigate the properties of isotropic conductive materials in electrostatic contexts
  • Learn about boundary value problems in partial differential equations
USEFUL FOR

Physicists, electrical engineers, and students studying electrostatics or partial differential equations will benefit from this discussion, particularly those interested in the behavior of electric potentials in conductive materials.

ZombieCat
Messages
4
Reaction score
0
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?
 
Physics news on Phys.org
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"...
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
Replies
11
Views
3K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K