# Resultant Capacitance of a capacitor having a combination of dielectrics

In a few questions, I have some confusion on how to take the resultant capacitance(Of a parallel plate capacitor) with a combination of dielectric in between.

Here is a an example

k1,k2,k3 represent the dielectric constants of the dielectric slabs. A is the plate area, d the distance between the plates.I know that we could split it into equivalent capacitors and from there it is easy to find the resultant capacitance.

But I am confused on how to split them. I have 2 possible approaches

(i)

(ii)

(k1,k2,k3 represent the dielectric constants of the dielectric slabs. A is the plate area, d the distance between the plates)

The 2nd part makes more sense to me and it also turn out to be the right method. However I cannot understand why the 1st part shouldn't work out.

Here is another example:

Even here I could split it in 2 ways-

(i)

(ii)

Here (i) is the correct way but (ii) isn't. I do not understand how either could be wrong or right.

Would really help if someone could point out what I am missing out.

(p.s: The spaces between the dielectrics that I have left in my solution are just to illustrate how I split the system. They indicate no real spaces.)

• In the first case (question from jee 2000), are you sure about the second way being the right answer? K1 and K3 in series as well as K2 and K3 in series, and then the combination in parallel seems correct. – Cluse Mar 7 at 9:50
• While dividing , don't disturb equipotential surfaces. – user280583 Mar 7 at 10:06
• @Cluse I used an unofficial answer key....it might be wrong – Vamsi Krishna Mar 7 at 10:48
• @MilanPaul Yeah I thought of that too..... however I had trouble identifying them. Correct me if I am wrong, a metal plate must be equipotential but the surface of a dielectric need not, right?(electrostatic condition) – Vamsi Krishna Mar 7 at 10:50

I think It all comes down to what is a parallel connection and what is a series connection. The potential over a capacitor is given as $$V=Ed$$ , where $$E$$ is the reduced electric field due to the insertion of dielectric k, as $$E=\frac{E_°}{k}$$ , where $$E_°$$ is the electric field with air as the dielectric medium. This implies that the electric field will be different for different dielectrics and hence the voltage will also change. Considering question from jee 2000, the voltage over $$k_2$$ and over $$k_1$$ won't be the same. Hence they are not in parallel. We are forced to divide the capacitor plate in such a way that each dielectric gets one half of $$k_3$$, or essentially they are in series. The same applies for the second question from 2015; we can't consider them in parallel and hence have to take them in series. After finding the equivalent series capacitance, we have two dielectrics now. But this time they are in parallel, because their end plates share a common potential. $$V=Ed$$ still holds, but the charges on the plates equalise the plate potential.