Electric field in a second dielectric given a 2 dielectric system

[willDavidson]

Homework Statement

The boundary between dielectric regions is defined by the plane $x+2y+3z=10$. The region containing the origin is refereed to as medium 1 $(x+2y+3z=10)$ and is assumed to have a permittivity $\epsilon_1=2\epsilon_0$. Medium 2 is assumed to be the free space of vacuum. The fields in both regions are static and uniform. If $E_1=2i+3j+4k$ $V/m$. Find E2. No surface charge is presented at the boundary.

Relevant Equations

$E_1 \epsilon_1=E_2 \epsilon_2$
$\oint_S E \cdot dl=0$
$\oint_S D \cdot ds$

I tried approaching this by finding the tangential and normal electric fields. Is this the correct approach? I've attached a drawing of the surface provided.

$\oint_S E \cdot dl=0$
$E_{tan1}\Delta x-E_{tan2}\Delta x=0$

We know that
$E_{tan1}=E_{tan2} Next, we can find the normal component using$$\oint_S D \cdot ds$
$D_{N1}\cdot dS-D_{N2}\cdot dS=Q$
$D_{N1}-D_{N2}= \frac Q {dS}$
$D_{N1}-D_{N2}=\sigma$

Since the problem defined no surface charge at the boundary
$D_{N1}-D_{N2}=0$
$D_{N1}=D_{N2}$

Now we use
$D=\epsilon E$
$E_1 \epsilon_1=E_2 \epsilon_2$
$E_2=\frac {\epsilon_1} {\epsilon_2}E_1$

Solution
$\epsilon_1=2\epsilon_0$
$E_2=\frac 1 2 E_1$
$E_1=2i+3j+4k V/m$
$E_2=\frac 1 2 (2i+3j+4k) V/m$
$E_2=i+\frac 3 2 j+2k$

 

[Gordianus]

May I suggest computing En1 and Et1 at the boundary?

 

[Gordianus]

I'm sorry. My mistake. I didn't realize the post was so old . The right panel presented it as without answer.