- #1
fluidistic
Gold Member
- 3,926
- 262
Homework Statement
I'm asked to find
1)The capacitance
2)The electric field
3)The surface charge density
4)The polarization charge at the interface between the dielectrics.
Of 2 set ups. The first one is a parallel plate capacitor which is half filled with a dielectric ##\varepsilon _1## and half filled with a dielectric ##\varepsilon _2##. The separation of the 2 dielectrics is a vertical plane.
In the other set up, the separation is a horizontal plane.
Homework Equations
C=Q/V.
##\vec D= \varepsilon \vec E##.
##(\vec D _2 - \vec D_1 )\cdot \hat n _{21}= \sigma {free}##.
##(\vec E_ 2 - \vec E_1 ) \times \hat n _{21}=0##.
##\sigma _{\text{pol}}=- (\vec P_2 - \vec P_1 ) \cdot \hat n_{21}## with ##P_i=(\varepsilon _i - \varepsilon _0) \vec E_i##.
The Attempt at a Solution
Strangely I'm asked to find the capacitance before the electric field?
So I go with the first set up. I assume a plate has a charge Q and the other -Q and that they are separated by a distance d.
I'm interested in the difference of potential between both plates, "V".
Now I believe that the D field will be different according the region of the dielectric. However ##\vec D_1 = \varepsilon _1 \vec E_1## where ##\vec E _1 = - \vec \nabla \Phi _1## for the first region and ##\vec D_2 = \varepsilon _2 \vec E_2## where ##\vec E _2 = - \vec \nabla \Phi _2## for the 2nd region. Now when I want to get "V", or in my terms ##\Phi (d)- \Phi (0)##, I realize that the result will depend upon which Phi I choose. In other words it seems that "V" depends on the region, which of course is totally false. I don't know what I'm doing wrong.
Last edited: