# Breakdown in parallel plate capacitor with dielectric

## Homework Statement

Two related questions (from Electromagnetics by Notaros)
The dielectric of a parallel-plate capacitor has two homogeneous parts, referred to
as dielectric 1 and dielectric 2. The boundary surface between the parts is flat and
parallel to the capacitor plates. The dielectric strength of dielectric 1 is twice that
of dielectric 2. If a voltage larger than the breakdown voltage of the capacitor is
applied across the capacitor terminals, the breakdown occurs in
(A) dielectric 1.
(B) dielectric 2.
(C) both dielectrics simultaneously.

and the second one:
Repeat the previous question but assuming that the boundary surface between
the two material parts constituting the dielectric of the parallel-plate capacitor is
perpendicular to the capacitor plates.

D=εE

## The Attempt at a Solution

In the first case, the thought the D field would be same in both dielectrics, because of the boundary conditions. Therefore, the dielectric with the smaller ε would reach breakdown first. In the second, the E fields are the same (tangential E field is equal across dielectric boundaries) and so the two would breakdown simultaneously (same E and same voltage).
However, the solution manual gives the first answer to be D) and the second to be B)
Does anyone know why this would be?

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TSny
Homework Helper
Gold Member
Hello, trotterlane. Welcome to PF!

I think that maybe the reason you are having difficulty with this question is that you are not clear on the definition of "dielectric strength". For example, what does it mean exactly to say that dielectric 1 has twice the dielectric strength of dielectric 2? Does that tell you anything about the dielectric constants ε of the two materials?

Otherwise, your reasoning about the magnitudes of D and E for the two cases appears to me to be correct.

ShayanJ
Gold Member
Your problem is that,break down voltage is not a property of the capacitor but a property of each dielectric and in that,the problem itself is wrong in saying " If a voltage larger than the breakdown voltage of the capacitor is
applied across the capacitor terminals".
Another point is that,the electric field is important because usually the break down field strength is given.
Assum the break down field of dielectric 1 is $E_1$ and that of dielectric 2 is $E_2$.Because as you said,D field is the same in both,the electric field in each of them becomes $\frac{D}{\varepsilon_1}$ and $\frac{D}{\varepsilon_2}$. If $E_1<E_2$,then when we have $\frac{D}{\varepsilon_1}=E_1$,dielectric1 will break down but not the capacitor itself because dielectric 2 is still a dielectric.The break down of dielectric1 only changes some of the properties of our capacitor.
But in the second question,we still have the electric fields by $\frac{D}{\varepsilon_1}$ and $\frac{D}{\varepsilon_2}$,and still if $E_1<E_2$,the break down occurs in the first dielectric but this time,the capacitor will break down because dielectric1 connects the two terminals and when it breaks down,i.e. becomes conducting,it discharges the capacitor.