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I wonder if the capacitance is the same if we have two plates with a dielectric material between. I mean, is it the same when the dielectric material only touches one of the plates, compared with when it touches both plates?

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No it isn't the same. After all, the electric field in the dielectric will be different (reduced). By placing an infinitesimally small metallic slab along the bottom face of the dielectric, you can model the capacitor as two capacitors in series, like in this diagram:enter image description here

Hope this helps.

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  • $\begingroup$ okay. So in this case $C_1>C_3>C_2$? If we assume the are separated. And what about if we have same capacitor as $C_1$ but instead of full dielectric material we have half of it, and it touches both sides. Is it then $\frac{fd}{2}$? Thanks! $\endgroup$
    – MathLover
    Oct 28 '21 at 13:02
  • $\begingroup$ @MathLover, What is $C_3$? If the dielectric occupies half of the space, then $fd = d/2$ and hence $f = 1/2$, I didn't understand your question. $\endgroup$
    – Cross
    Oct 28 '21 at 13:09
  • $\begingroup$ $C_3$ is the capacitor in the left in your picture. My main question is when is the capacitans as big as possible? I see in your picture that $C_1$ is the biggest and $C_2$ the smallest. But, how about other cases? $\endgroup$
    – MathLover
    Oct 28 '21 at 13:54
  • $\begingroup$ The picture on the left and right are equivalent. You can split the entire capacitance (left) into two capacitors $C_1$ and $C_2$, which are in series as shown on the picture in the right. The naming $C_1$ and $C_2$ are arbitrary, and $fd$ just represents a fraction f of the value $d$ (which represents the separation between the two plates of the capacitor). Maximising the capacitance is naturally when $fd = d$; that is, the dielectric occupies the entire space. You can prove it mathematically by setting $\frac{dC_{eq}}{df}$ to zero, where $C_{eq}$ is the equivalent capacitance of the system. $\endgroup$
    – Cross
    Oct 28 '21 at 14:04
  • $\begingroup$ I understand now! Thanks a lot!! $\endgroup$
    – MathLover
    Oct 28 '21 at 14:26

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