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Alettix
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Homework Statement
a) State the boundary conditions for the electric field strength E and electric flux density D at a planar interface separating two media with dielectric constants ε1 and ε2.
b) A parallel plate capacitor with a plate separation d is filled with two layers of different materials which have dielectric constants ε1 and ε2, conductivities σ1 and σ2 and layer thicknesses d1 and d2, respectively. The two materials completely fill the region between the plates so that d1 + d2 = d. A constant voltage V is applied between the plates. By considering the voltage drop across the capacitor and the current flow though the capacitor, find expressions for the magnitudes of the electric fields in both materials.
c) Derive an expression for the current density flowing through the capacitor.
d) Find expressions for the total surface charge density on the interface between the two materials and for the free surface charge density on the interface between the two materials.
Homework Equations
Ohm's law: ##\vec{J} = \sigma \vec{E} ##
Capactiance: ##C = \frac{\epsilon \epsilon_0 A}{d} ##
Electric field intensity: ##D = E \epsilon \epsilon_0 ##
The Attempt at a Solution
a) The boundary conditions are:
- Perpendicular component of D continuous → perpendicular E discontinuous
- Parallel component of E continuous → parallel component of D discontinuous
However, in the derivation of the first condition, it is assumed that there is no free charge on the boundary so my first question is: are these equations valid in this case?
b) Using ##V = E_1d_1 + E_2d_2 ## and conservation of current density: ##\sigma_1 E_1 = \sigma_2 E_2## we find:
## E_1 = \frac{\sigma_2 V}{\sigma_1d_2 + \sigma_2d_1} ##
## E_2 = \frac{\sigma_1 V}{\sigma_1d_2 + \sigma_2d_1} ##
I also tried using current conservation with ##I = \frac{V_n}{C_n\omega}##, giving expressions involving ##\epsilon_n## instead of ##\sigma_n##, but I am not sure if the capactance formula is valid in this case.
c) The current density is hence: ##J = \frac{\sigma_1\sigma_2 V}{\sigma_1d_2 + \sigma_2d_1} ##
d) From this point on I am a bit really I think we have:
##\nabla \cdot E = \frac{\rho_{total}}{\epsilon_0}## and ## \nabla \cdot E = \frac{\rho_{free}}{\epsilon_0\epsilon}## from Gauss's law.
The first relation is easy to apply to find the total charge, but what about the free charge, which ##\epsilon## ought to be used?
Many thanks!
UPDATE: It goes under the integral - doesn't it? Like ##E_1\epsilon_0\epsilon_1 - E_2\epsilon_0\epsilon_2 = \rho_{free} ##?
However, I am still interested in finding out if the E-D boundary conditions are applicable in conducting media.
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