Inserting a conductor in a parallel-plate capacitor

AI Thread Summary
The discussion focuses on the equivalent capacitance of a parallel-plate capacitor with a conducting plate inserted between the plates. When the conductor is partially inserted (0 < x < l), the configuration can be represented by three capacitors: C1, C2, and C3, each defined by their respective dimensions and positions. There is confusion regarding the interpretation of C3, which is clarified to represent the capacitance of the space to the right of the conductor rather than the conductor itself. When the conductor is fully inserted (l < x < L), the arrangement changes to include four capacitors, indicating a more complex interaction. Understanding these configurations is crucial for accurately calculating the overall capacitance.
Joshua Benabou
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##.

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:
uKSaN.jpg


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:

34tCi.jpg


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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Hi,

As Sherlock Holmes would say: then the logical conclusion is that your guess was wrong ...

Look at ##C_3##: is your guess logical if e.g. ##x=0## ?
 
Joshua Benabou said:
C3C3C_3 the capacitor formed by the conductor itself
As BvU has suggested, the issue is with your interpretation of ##C_{3}##. The capacitor ##C_{3}## is considered to be the capacitance of the space to the right of the conductor, leading to the form of ##C_{3}## given in your notes.
 
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