# Maximum stored energy in a capacitor with different materials?

In class today, we saw dielectrics and their application to capacitors (e.g. parallel plate, cylindrical, spherical capacitors). A question came up which discussed how when we have a capacitor that had several dielectrics inside, then to calculate the maximum energy stored we'd have to calculate the minimum potential difference for which one of the materials break. This is kind of hard for me to understand, where can I read on about this concept?

• Each material has a voltage point at which the material transitions in another state which usually is more conductive. This is the breaking point you're looking at. I'd suggest you to look at which materials you have and look for breaking point voltages required with Google. Jan 17, 2018 at 11:34
• Thank you. I understand the concept of breakdown electric field and voltages for a single material, what I am confused with is how to work with several materials that have different breakdown fields, as usually, when material 2 reaches this value, material 1 for instance might have already broken down.
– Bee
Jan 17, 2018 at 12:33
• You have to look at the weakest link. Also, the electrical field changes depending on the material. The material which has the lowest dielectrical constant also has the biggest electrical field over it (relative to it's distance from the source) for series connection of dielectrics. You have to calculate for that material. Jan 17, 2018 at 12:55
• Bee, in an added note to my answer below, I also gave the the maximum stored energy in the capacitor. Jan 18, 2018 at 17:31

Usually, a dielectric has a specific critical breakdown field field $E_{br}$. Thus, when you have a parallel plate capacitor with layers of different dielectrics, you get breakdown in one of the layers when its critical field is reached. The electric field $$E=V/d$$ in the capacitor with plate distance $d$ is independent of the dielectric layer considered. Therefore you get a breakdown first in the layer with the lowest $E_{br}$ when $$E=V/d \ge E_{br}$$
Added note: Once you know the voltage $V_{br}$ at which this weakest dielectric breaks down, you have also the the maximum stored energy in the capacitor $$U_{max}= C V_{br}^2/2$$