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So here I have a capacitor with two plates , one positively charged and one negatively, as shown in the figure. Now the material between them is $k_1$ for thickness d/2 and $k_2$ for the other d/2. There is no plate or anything between the separation. We have to find the equivalent capacitance.

What my school teacher did was that she took them in series and drew them as two capacitors with one which has medium $k_1$ in between and the other with $k_2$ in between. But I think this is wrong. I took it as a capacitor which is in a medium $k_1$ and has a slab in between of medium $k_2$. I know that if there is a capacitor with distance between the plates as d and has air between it and has a slab in between the plates such that it's thickness is t and it's dielectric constant is k then it's equivalent capacitance is $\frac{\epsilon A}{d-t(1-\frac1k)}$.

So if I imagine it as the scenario above with dielectric constant $k_1$ in the back and a slab of dielectric $k_2$, then we just put $t=d/2$,$k=k_2$ and multiply the above by $k_1$, which according to me should be the answer but if I go by my teachers method and do the math, we get $\frac{k_1 k_2 \epsilon A}{(k_1 + k_2)d}$.

Who is wrong here?

EDIT: Here A is the area of the plate.

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Your teacher is right. Although you missed a factor of $2$ in last line.

The case you are saying is NOT equivalent to this one. How do you plan on filling dielectric $K_1$ on the lower half which is already filled with $K_2$? This makes no sense.

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  • $\begingroup$ No no no. I'm not saying that. I'm saying that I have derived a formula for the eq. capacitance when a slab of dielectric k is introduced in between the plates already filled with air. I'm comparing it to when the plates are already filled in between with $k_1$ and a slab of $k_2$ is introduced. $\endgroup$ – Rohinb97 Apr 18 '14 at 18:30
  • $\begingroup$ But here initially air was there and then $K_1$ and $K_2$ were introduced independently $\endgroup$ – evil999man Apr 18 '14 at 18:44
  • $\begingroup$ I took $k_1$ initially and then added it. Cause if you derive with $k_1$ initially, you will get the thing I mentioned above. And btw, even if my teacher is right, how can she make two plates 4? Like if you draw it in series as my teacher and you are saying, from where does the extra plate(s) come from? $\endgroup$ – Rohinb97 Apr 20 '14 at 6:03
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    $\begingroup$ @Rohinb97 Putting a thin metal in between wont change capacitance. $\endgroup$ – evil999man Apr 20 '14 at 12:16
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I don't really follow your method, but it's probably sound because it gives the same answer as your teacher's method, when we put $t=\frac{d}{2}$, $k_1=k$, $k_2=1$, and we don't lose the factor of 2, as pointed out by evil999man.

Your teacher's method is the standard way of doing this question. As already pointed out, the introduction of a thin conducting plate between the dielectrics doesn't change the capacitance, as this plate will have equal and opposite charges induced on its two faces, but it shows how the original problem can be reduced to a simple matter of capacitors in series.

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