Calculating Capacitance With Dielectric

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To calculate the new capacitance of a capacitor with a dielectric inserted between its plates, treat the system as two capacitors: one with the dielectric and one without. When a dielectric does not touch the plates, it can be considered as three capacitors by introducing thin metal plates at the boundaries. The capacitance can be calculated using the series formula for capacitors. The formula for capacitance with a dielectric is KAE0 / Kd-x(K-1), where K is the dielectric constant, A is the area, and d is the distance between plates. When dealing with multiple dielectrics, divide the system into individual capacitors and calculate their total capacitance in series.
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Lets say There is a capacitor with the are of Plates A and the distance between the plates say is 5m if a dielectric of similar dimensions except the width is 3m is filled between the Capacitor plates how to calculate the new Capacitance ??
 
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Imagine a very thin metal plate at the boundary of the two dielectrics. Now you have two capacitors, one with the dielectric and one without. Simply calculate the value of the two capacitors and use the capacitors in series formula to arrive at your answer.
 


Okay so let's say the capacitor plates are at a distance d from each other and the inserted dielectric is of x width and constant K the formula comes out to be KAE0 /Kd-x(K-1) which solves the problem thanks ! What is confusing me is if the dielectric is displaced towards one direction that is it does not touch any of the plates will it still be considered as two capacitors ?
 


Okay, so the dielectric is spaced off both plates. Just imagine a very thin metal plate on both sides of the dielectric, now we have three capacitors, calculate each one then calculate the total value for those three capacitors in series. Bottom line is when you encounter a boundary, just imagine a metal plate there splitting the capacitor in two. So if you have a problem where different dielectrics are stacked just divide the problem into so many individual capacitors, then figure them in the series circuit.
 
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