Consider a parallel plate capacitor formed by two plates of length ##L## and width ##d##, separated by a distance ##e##. There is a vacuum in between the plates. Let's note the capacitance of this arrangement ##C_0##.(adsbygoogle = window.adsbygoogle || []).push({});

I insert a conducting plate of length ##l=L/2##, with ##D##, and thickness ##e' <<e##. The position of the plate is measured by its ##(x,y)## coordinates, as shown below:

I would like to find the equivalent capacitance of this apparatus in terms of the distance ##x##.

Of course if ##x<0##, the conductor is not inserted at all so the capacitance remains unchanged, ##C_0##.

Consider the case where the conductor is inserted partially, i.e ##0<x<l##.

According to my notes, in this case the apparatus is equivalent to the arrangement of capacitors below:

where

##C_1=\frac{\epsilon_0Dx}{e-y-e'}##

##C_2=\frac{\epsilon_0Dx}{y}##

##C_3=\frac{\epsilon_0D(L-x)}{e}##

**I do not understand why this configuration is equivalent to the arrangement of capacitors given above.**

I guess ##C_1## is the capacitor formed by the top plate and the conductor, ##C_2## the capacitor formed by the bottom plate and the conductor, and ##C_3## the capacitor formed by the conductor itself. However this leaves me confused as the capacitance for the conductor should then be:

##C_3=\frac{\epsilon_0Dx}{e}##

Finally, if we now consider the case where the conductor is fully inserted, i.e ##l<x<L##, then apparently the capacitor arrangement changes completely and we now actually have four capacitors (2 in series, which are parallel with the other two). I don't understand why.

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# I Inserting a conductor in a parallel-plate capacitor

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