- #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##
