- #1

willDavidson

- 50

- 6

- 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##

##\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##