# Tricky: Inserting a conductor in a parallel-plate capacitor

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: 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.

## 2 Answers

In your configuration, $C_3$ is given by the electrodes at distance $e$ but the width is now $L-x$. Or put in another way, you can see you capacitor as the parallel of two capacitors, let's call $C_3$ and $C'_3$. Now, as you've figured out, $C'_3$ is given by $C_1$ and $C_2$ connected in series, and $C_3$ is like you original capacitor (without the conducting plate) but with electrodes with width $L-x$. So you have to consider the part of you electrodes that "see each other" When the conducting plate is totally inserted the situation is the same, you have part of the original electrodes that "see each other" both on the left and right of the conducting plate. Hope this helps.

• This is not well explained. Why do you consider the part of the electrodes which see each other? – Joshua Benabou Nov 6 '17 at 11:59
• In your system you have 3 conductors, let's say A(=top electrode), B(bottom electrode) and C(=conducting plate). With 3 conductors you have 3 capacitances: $C_{AC}=C_1$, $C_{BC}=C_2$ and $C_{AB}=C_3$ when the conducting plate is inserted. If the conducting plate is far away you'll have only $C_{AB}$. – Marco81 Nov 6 '17 at 22:13

Here I draw a simple illustration to show the configuration of the given system. ANOTHER CONFIGURATION 