- #1
Phillipe
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Homework Statement
We have two infinite conductor plates, with two dielectrics (with permitivities K1 and K2) put together between them so that the capacitor may be taken as two capacitors connected in series. We know that the maximum electric field for the first dielectric is Er1, and we also know the thicknesses of both dielectric layers (we'll call the thicknesses d1 and d2).
A) What's the value of the maximum electric field for the second dielectric, so that both will break at the same time?
B) Which voltage do we need to apply in order to get such field?
(When I say "maximum electric field", I'm referring to the electric field required to break the dielectric and turn it into a conductor)
Homework Equations
Let E0 be the electrical field applied to the capacitor, and Vr be the required voltage to break both dielectrics.
[tex]{E_r} = \frac{{{E_0}}}{{{k_i}}}[/tex] (with i=1,2)
[tex]V = Ed[/tex]
The Attempt at a Solution
Since
[tex]{E_0} = {E_{r1}}{k_1}[/tex]
and
[tex]{E_0} = {E_{r2}}{k_2}[/tex]
We can write
[tex]{E_{r1}}{k_1} = {E_{r2}}{k_2}[/tex]
therefore
[tex]\frac{{{E_{r1}}{k_1}}}{{{k_2}}} = {E_{r2}}[/tex]
which gives us the required field to break the second dielectric. I assume that if we apply an electric field of at least E0 we'll break them both. Which leads us to calculate the required voltage:
[tex]{V_r} = {V_1} + {V_2}[/tex] (consider both dielectrics as two capacitors connected in series)
threrefore, the required voltage would be
[tex]{V_r} = {E_{r1}}{d_1} + {E_{r2}}{d_2}[/tex]Is my procedure correct? Thanks a lot.