- #1

C12H17

- 4

- 0

## 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]

_{0}εE

_{x}+[itex]\vec{y}[/itex]

_{0}εE

_{y}+[itex]\vec{z}[/itex]

_{0}εE

_{z}

D

_{x}=εE

_{x}

D

_{y}=εE

_{y}

D

_{z}=εE

_{z}

then

ρ=d Asin(x)cos(y)E

_{x}/ dx + dAsin(x)cos(y)E

_{y}/dy + d Asin(x)cos(y)E

_{z}/ 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}/dx

^{2}+Acos(x)cos(y) d [itex]\varphi[/itex]

^{2}/ dy

^{2}+ Asin(x)cos(y) d [itex]\varphi[/itex]

^{2}/ dz

^{2}

and now i don't know. ;D

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