My question is how would you go about calculating the capacitance of a parallel plate capacitor which has two mediums in between the parallel plates. I am aware that for a regular parallel plate capacitor you can calculate its capacitance with the following equation:

\begin{equation} C=\varepsilon\frac{A}{d} \end{equation}

However, for a capacitor as the one shown in the following image, this equation is not applicable because there are multiple dielectrics.

Parallel plate capacitor with multiple dielectrics

My guess is that you could do something like this (assuming both dielectrics are equal volume):

\begin{equation} C=\frac{\varepsilon_1\varepsilon_2}{2}\frac{A}{d} \end{equation}

Is this correct? otherwise, what would be the correct way of doing this?


closed as off-topic by Ben Crowell, ZeroTheHero, Jon Custer, Chair, Kyle Kanos Feb 12 at 11:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Ben Crowell, ZeroTheHero, Jon Custer, Chair, Kyle Kanos
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ What are your thoughts? What is the basic definition capacitance? What happens to voltage when we stack one dielectric wafer on another? $\endgroup$ – exp ikx Feb 10 at 11:21
  • $\begingroup$ What is your background knowledge? Do you know Gauss' theorem? Are you familiar with the vector $\vec{D}$? $\endgroup$ – FGSUZ Feb 10 at 13:54
  • 1
    $\begingroup$ I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. $\endgroup$ – Ben Crowell Feb 10 at 16:59

imagine a plate between two dielectric materials, now this plates seperates two dielectric and we can assume them as two seperate parallel plate capacitors connected in series

equivalent capacitance is $1/c=1/c1+1/c2$


Not the answer you're looking for? Browse other questions tagged or ask your own question.