Electrodynamics: Electrostatic field potencial in Cartesian coordinates

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SUMMARY

The discussion focuses on determining the electrostatic field potential function \(\varphi\) in Cartesian coordinates, given the absolute permittivity function \(\epsilon(x, y, z) = A \sin(x) \cos(y)\). The relationship \(\vec{D} = \epsilon \vec{E}\) and \(\vec{E} = -\nabla \varphi\) is established, leading to the divergence equation \(\nabla \cdot \vec{D} = \rho\). The user attempts to derive the charge density \(\rho\) but encounters difficulties due to the absence of charge distribution and boundary conditions.

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  • Understanding of electrostatics and Maxwell's equations
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Homework Statement



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

Homework Equations



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

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 don't know. ;D
 
Last edited:
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You're not given any charge distribution, any boundary conditions?
 

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