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Electrodynamics: Electrostatic field potencial in Cartesian coordinates

  1. Jan 23, 2013 #1
    1. The problem statement, all variables and given/known data

    It's given that absolute permitivity is a coordinate function: ε (x, y, z) = Asin(x)cos(y), where A=const

    2. Relevant equations

    We need to find an electrostatic field potential function [itex]\varphi[/itex] in Cartesian coordinate system.

    3. The attempt at a solution

    I tired to solve, but I don't know if it's ok. Check, please?

    [itex]\vec{D}[/itex]=ε[itex]\vec{E}[/itex] and [itex]\vec{E}[/itex]= - grad[itex]\varphi[/itex]
    div[itex]\vec{D}[/itex]=ρ
    ρ=[itex]\frac{dDx}{dx}[/itex]+[itex]\frac{dDy}{dy}[/itex]+[itex]\frac{dDz}{dz}[/itex]
    [itex]\vec{E}[/itex]=[itex]\vec{x}[/itex]0εEx+[itex]\vec{y}[/itex]0εEy+[itex]\vec{z}[/itex]0εEz
    Dx=εEx
    Dy=εEy
    Dz=εEz
    then
    ρ=d Asin(x)cos(y)E x / dx + dAsin(x)cos(y)E y /dy + d Asin(x)cos(y)Ez / dz=Asin(x)sin(y)[itex]\frac{dE}{dx}[/itex]+Acos(x)cos(y)[itex]\frac{dE}{dy}[/itex]+Asin(x)cos(y)[itex]\frac{dE}{dz}[/itex]=Asin(x)sin(y) d [itex]\varphi[/itex]2 /dx2 +Acos(x)cos(y) d [itex]\varphi[/itex]2 / dy2 + Asin(x)cos(y) d [itex]\varphi[/itex] 2/ dz2

    and now i dont know. ;D
     
    Last edited: Jan 23, 2013
  2. jcsd
  3. Jan 27, 2013 #2

    rude man

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    Homework Helper
    Gold Member

    You're not given any charge distribution, any boundary conditions?
     
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