How Does Inserting a Copper Slab Affect a Capacitor's Performance?

In summary, when a slab of copper is inserted into a parallel plate capacitor, the total capacitance of the system becomes the sum of the capacitance of the two capacitors in series. The ratio of the stored energy before and after the insertion can be found by dividing the energies of the two capacitors, which shows that the work done on the slab is positive and therefore it is pushed in. The work done by the battery to maintain a constant voltage across the capacitor contributes to the change in energy. The relationship between potential difference and electric field can be found by using the knowledge of the electric field in the dielectric and free space regions. The book "Electromagnetic Theory" by Reitz is recommended for further understanding of dielectric sl
  • #1
stunner5000pt
1,461
2
A slab of copper is thrust into parallel plate capacitor as shown in the figure.
a) What is the capacitance after the slab is introduced?

This is like two capacitors in series so
[tex] C_{1} = \frac{\epsilon_{0} A}{d-b-x} [/tex]
[tex] C_{2} = \frac{\epsilon_{0} A}{x} [/tex]
when added it yields [tex] \frac{epsilon_{0} A}{d-b} [/tex]
Find the ratio of the stored energy before and after the slab is inserted if the voltage is kept constant
dividing U1 and U2 which are
[tex] U_{1} = \frac{1}{2} \frac{\epsilon_{0} A}{d-b} \Delta V^2 [/tex]
[tex] U_{2} = \frac{1}{2} \frac{\epsilon_{0} A}{d} \Delta V^2 [/tex]
[tex] \frac{U_{1}}{U_{2}} = \frac{d}{d-b} [/tex]
Find the work done on the slab as it is inserted. Is it pulled in or pushed in?
Well find the difference in the energy U1 - U2 right?
[tex] U_{f} - U_{i} = \epsilon_{0} A (\frac{1}{d-b} - \frac{1}{d} [/tex]
this change is positive value. So the slab must be pushed in. Is this right?
Please do advise on any mistakes i may have made.
Thank you for your help!
 

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  • #2
can someone advise me on whether i am right or not on the last part of this problem? I know the first and second parts are correct but what about the last part ? Since the nergy of the system goes up there must be external work done... so the slab mustb e pushed in, yes?
 
  • #3
stunner5000pt said:
can someone advise me on whether i am right or not on the last part of this problem? I know the first and second parts are correct but what about the last part ? Since the nergy of the system goes up there must be external work done... so the slab mustb e pushed in, yes?
Some of the external work is done by the battery or other source that keeps the voltage constant across the capacitor.
As the final capacitance is higher than the initial one and the voltage stays the same, the charge increases. The work done by the battery is the constant voltage multiplied by the increment of charge. The change of energy is the summ of your work and the work done by the battery
[tex] \Delta E = \Delta Q U + W [/tex]
ehild
 
  • #4
I have the same problem but how do i find the potential difference as a function of x. Also what is the electrostatic energy of the system and force on the dielectric.
 
  • #5
Do you know the electric field in the dielectric and in the free space region? What is the relationship between V and E ?
 
  • #6
Electromagnetic Theory by Reitz is a good book to learn more about dielectric slabs from...
 

What is a dielectric in a capacitor?

A dielectric in a capacitor is a material placed between the two plates of a capacitor, which helps increase the capacitance by reducing the electric field between the plates.

What is the role of a dielectric in a capacitor?

The main role of a dielectric in a capacitor is to increase the capacitance of the capacitor by storing more electrical energy. It also helps to prevent breakdown of the capacitor by reducing the electric field between the plates.

What are some common materials used as dielectrics in capacitors?

Some common materials used as dielectrics in capacitors include air, paper, mica, ceramic, and plastic. The choice of dielectric material depends on factors such as cost, dielectric strength, and temperature stability.

How does the dielectric constant of a material affect the capacitance of a capacitor?

The dielectric constant, also known as relative permittivity, is a measure of how well a material can store electrical energy. The higher the dielectric constant, the greater the capacitance of the capacitor.

Can the dielectric in a capacitor affect its performance over time?

Yes, the dielectric in a capacitor can affect its performance over time. Some dielectric materials may degrade or break down over time, leading to a decrease in capacitance and potentially causing the capacitor to fail. Also, changes in temperature, humidity, and other environmental factors can affect the dielectric properties of the material and therefore impact the performance of the capacitor.

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