# Is this combination of dielectric in series or parallel? [closed]

In this question, I was required to find the effective dielectric constant of the system. (Source:- JEE 2000). What I did was prove that the electric fields in K1 and K2 are having the same E fields inside them(as shown in the image ). I then concluded that since the fields are uniform, constant and the distance which the dielectrics k1 and k2 occupies are the same, the potential drop across them should be the same. ( I had asked a very similar question where the charges got redistributed for dielectrics placed similarly resulting in equal fields in them Why is the general definition of electric fields in dielectrics breaking down here? here.)

I hence concluded that Slabs k1 and k2 are in parallel and their equivalent would be in series with slab k3 and got an answer as below which was wrong.

The correct answer given was D which is obtained when we take the slabs k1 and k2 each to be in series with half of k3 each and then taken in parallel. The Working has been shown below:

Concerns:-

1. If the voltage differences across the slabs k1 and k2 are equal why are they in series with k3 and not in parallel with each other?
2. On top of this, the capacitors in series have the same charges on them which doesn't seem to be the case here as $$Q_{ind}$$= $$Q(1-\frac{1}{k})$$ and since k is different, the charges on all the slabs are different. One of my friends responded to this by stating that capacitors in series need not always have the same charge and gave an example of two capacitors connected by an inductor.

Possible Resolutions and theories:-

1. Some people pointed out to me that the fringing fields would potentially affect the potential difference being equal for k1 and k2
2. I theorized with one of my friends that the charge distribution of the metal plates would also change as pointed out in Why is the general definition of electric fields in dielectrics breaking down here? , which may potentially cause the slabs to have the same charge and hence be in series.(Although the potential drops across them are weirdly always equal>) Also note here that the charge distribution on the metal plates will be non uniform
• Who says the voltage across the slabs is equal? Also, this post is hard to read with all the handwritten calculations. If you're going to include them, you should use MathJax, but I don't actually think they're relevant at all to the question. – d_b Jun 19 '20 at 18:14
• I proved them to be equal as E fields in both are the same and their distances are same then by V=Ed both have same potential difference – Schwarz Kugelblitz Jun 19 '20 at 18:31
• First of all, your "proof" assumes that the electric field in each dielectric is uniform, which I don't think is true in this example. But I was responding to your question "If the voltage differences across the slabs are equal why are they in series?" I thought you were asking about 1 and 3 or 2 and 3, since these are the ones in series. If you're asking about 1 and 2, the answer is: they're not in series. – d_b Jun 19 '20 at 18:40
• I am saying is K1 in parallel with k2.Please refer to my diagrams – Schwarz Kugelblitz Jun 19 '20 at 18:42
• And why would the fields not be uniform ? – Schwarz Kugelblitz Jun 19 '20 at 18:43

But what's wrong with first taking them in parallel and then series

Let the three capacitances in your proposed connection be $$C_1$$, $$C_2$$, and $$2C_3$$. The equivalent capacitance is then

$$C = 2\frac{(C_1 + C_2)C_3}{C_1 + C_2 + 2C_3}$$

Now, think about what happens when, say, $$k_1$$ is made much, much greater than $$k_2$$ and $$k_3$$. The equivalent capacitance would go to

$$C\rightarrow 2\frac{C_1C_3}{C_1} = 2C_3 = 2\frac{\epsilon_0k_3A}{d}$$

This doesn't pass the sanity check. It doesn't seem plausible that making $$k_1$$ arbitrarily large should negate the effect of the $$k_2$$ slab entirely.

The equivalent capacitance of the correct parallel combination of two series connected capacitors is

$$C = \frac{C_1C_3}{C_1 + C_3}+\frac{C_2C_3}{C_2 + C_3}$$

and the limit of arbitrarily large $$k_1$$ is

$$C\rightarrow C_3 + \frac{C_2C_3}{C_2 + C_3} \lt 2C_3$$

This does pass the sanity check. Only if $$k_2$$ is also made arbitrarily large does the equivalent capacitance approach $$2C_3$$.

So where has your thinking gone wrong? If the connection you propose is valid, then we can insert an ideal conductor plate at the interface of the $$k_3$$ dielectric with the others without changing the system. But I'm almost certain that's not the case.

With the plate there, the electric field lines must be 'vertical' (normal to the plate) at the interface. But, without the plate there, I don't believe that there's a requirement that the field lines are normal to the dielectric interfaces.

• Umm but aren't the field lines already normal?I still dont quite follow how that causes a problem ? Does my proof of E1=E2 only hold for infintely large plates – Schwarz Kugelblitz Jun 19 '20 at 21:47
• @SchwarzKugelblitz, I just discovered that your question is essentially an exact duplicate of this one. I think it would be good to try to merge the two together somehow. – Alfred Centauri Jun 19 '20 at 22:40
• But I proved that the middle interface is an equipotrntial when I proved that the fields in those dielctrics were the same? – Schwarz Kugelblitz Jun 19 '20 at 22:42
• Wait if they are in series how do they have same charges ? – Schwarz Kugelblitz Jun 19 '20 at 22:43
• @SchwarzKugelblitz, two series connected capacitors have the same voltage across only if they have the same capacitance. Is that what you're asking? – Alfred Centauri Jun 19 '20 at 22:46