Inserting a conductor in a parallel-plate capacitor

This leads to the equivalent arrangement of capacitors shown in the diagram. In summary, the equivalent capacitance of the arrangement can be found by considering the individual capacitances of each component, including the space to the right of the conductor, and combining them according to the rules for capacitors in series and parallel.
  • #1
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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  • #2
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## ?
 
  • #3
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.
 

1. What is a parallel-plate capacitor?

A parallel-plate capacitor is a type of electronic component that stores electrical energy by creating an electric field between two parallel plates. It consists of two conductive plates separated by an insulating material, known as the dielectric.

2. Why would I want to insert a conductor in a parallel-plate capacitor?

Inserting a conductor in a parallel-plate capacitor can change the capacitance of the capacitor. This can be useful in applications where you need to adjust the amount of electrical energy stored in the capacitor.

3. How does inserting a conductor affect the electric field in a parallel-plate capacitor?

Inserting a conductor between the plates of a parallel-plate capacitor will change the distribution of the electric field between the plates. This is because the conductor creates a new path for the electric field to flow through, altering the overall capacitance of the capacitor.

4. What factors affect the capacitance of a parallel-plate capacitor with a conductor?

The capacitance of a parallel-plate capacitor with a conductor is affected by the distance between the plates, the size and shape of the plates, the type of dielectric material used, and the presence of any other conductors or materials nearby that may also affect the electric field.

5. Can inserting a conductor damage a parallel-plate capacitor?

In most cases, inserting a conductor into a parallel-plate capacitor will not damage the capacitor. However, if the inserted conductor is not properly insulated and comes into direct contact with the plates or other conductive materials, it can cause a short circuit and potentially damage the capacitor.

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