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Consider a parallel plate capacitor of area $A$ with a distance between the plates $d$, disconnected from the battery. I wonder, how would the capacitance of the system change if we placed a conductor or an insulator (which also have some width $l$ < $d$) of smaller area $A'$ between the plates?

Apparently, we can't neglect edge effects in such scenario, so it is not clear how to calculate voltage between the plates. This system is equivalent to two capacitors in series (1-3 and 4-2), however, both of these capacitors seem to have different charges on each plate. In case of conductor, since the field inside is zero, charge on each side must be $Qconductor=Qplates*A'/A$, and in case of insulator, since the field inside must be decreased by $\epsilon$ - $Qinsulator=Qplates*A'*(1-1/(\epsilon))/A$. So we get two capacitors of different surface areas and with different charges. How can I calculate the capacitance of any of them? Is there another way to deal with the initial problem?
image

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closed as off-topic by ACuriousMind, Sebastian Riese, user36790, Kyle Kanos, CuriousOne Feb 25 '16 at 7:01

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You have two capacitors in parallel - represented by the two area and one of those capacitors is actually two capacitors in series.
Those two series capacitors are either one with one type dielectric and one with another dielectric or if it is a conductor which is placed between the plates then just two capacitors with the same dielectric.

So a pair of capacitors in series in parallel with another capacitor.

Later

enter image description here

If the shaded bit is a conductor then there is no capacitor there just a conducting wire.
You can move the shaded area to any convenient spot. So in my diagram it goes top right.

Later still
enter image description here

Hopefully this shows that order does not matter.

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  • $\begingroup$ I've added a picture. Could you tell which conductors are parallel/in series, because I'm quite lost? $\endgroup$ – Dmst Feb 24 '16 at 13:11
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    $\begingroup$ @Dmst I have added a diagram. I hope that it helps? $\endgroup$ – Farcher Feb 24 '16 at 15:17
  • $\begingroup$ Thanks, I was able to get the correct answer, but I still can't figure out, why we may move the conductor/insulator to wherever we want. Could you explain it? $\endgroup$ – Dmst Feb 24 '16 at 15:49
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    $\begingroup$ @DMst I have added another diagram to show you that for the series arrangement the order in which you place the two dielectrics does not matter for the series part of the arrangement. You can do a similar analysis for the parallel part. $\endgroup$ – Farcher Feb 24 '16 at 16:03
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    $\begingroup$ In these calculations the edge effect is neglected to make life easier. The change due the edge effect can be made small by making sure that the separation of the plates is much less than the smallest linear dimension which makes up the area of the plates. So the gap might be 1 mm and the plates 300 mm x 300 mm. $\endgroup$ – Farcher Feb 24 '16 at 16:42
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yes, According to the formula of parallel capacitor.

Capacitance = k x ε0 x area / separation Where k = relative permittivity of the dielectric material between the plates. and we know that

C = q/V

equal both the equation you will get potential and charge. please do specific in your question next time with a picture of your problem. Arrange all the equation in proper manner.

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  • $\begingroup$ The first formula can't be used in this case $\endgroup$ – Dmst Feb 24 '16 at 13:14
  • $\begingroup$ Welcome to Physics! Note that this site has MathJax enabled, which means you can use Latex-like syntax to add in equations for readability. $\endgroup$ – Kyle Kanos Feb 25 '16 at 1:10

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