Parrallel Plate Capacitor Model Question

In summary, the parallel plates for the capacitor do not have to be identical, but it is recommended for a higher accuracy. The dimensions of the plates should be as close as possible to each other for the equations to be more accurate. The thickness of the plates should not be taken into account. The equations for capacitance should be used in MKS units, even if the given units are in inches. And the reason for the 4*pi in the second equation is to omit the constant ε0.
  • #1
UsualAnalyst
1
0
Hello, i am new to the forums here so forgive me if I have mistakenly started a matching thread that already exists.

So i have a task for building a parallel plate capacitor for two aluminum plates connected to a battery on opposite terminals. The target capacitance is 2000 microfarads. with a distance of 0.062 inches between the plates. Oh yea, the units of length are inches.

Since the units for the general equations for capacitance are in either CGS or MKS units, this leaves me confused. So, i have a few questions before completing the experiment.

Questions:
1) Do the parallel plates have to be identical; meaning that they have to be of the same material?
2) For the dimensions of the plates, do they have to be absolutely the same? meaning that can one be smaller or larger than the other?
3) When finding the area of the plates, are the thicknesses of the plates taken into account. (I know Area is L * W, and adding the thickness would mean Volume; however, since voltage is flowing through the plates, i wonder about it.)
4) Since the units of length that i am obligated to use are in inches would that interfere with the equations that are given for Capacitance such as:

C = εrε0[itex]\frac{A}{d}[/itex]

C = εr[itex]\frac{A}{4*Pi*d}[/itex]


5) One final question, for the equation would be why is the 4*pi applied in the second equation or is it used to omit ε0 constant?

Please and thank you in advance for anyone's time.
 
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  • #2
UsualAnalyst said:
Hello, i am new to the forums here so forgive me if I have mistakenly started a matching thread that already exists.

So i have a task for building a parallel plate capacitor for two aluminum plates connected to a battery on opposite terminals. The target capacitance is 2000 microfarads. with a distance of 0.062 inches between the plates. Oh yea, the units of length are inches.

Since the units for the general equations for capacitance are in either CGS or MKS units, this leaves me confused. So, i have a few questions before completing the experiment.

Questions:
1) Do the parallel plates have to be identical; meaning that they have to be of the same material?
2) For the dimensions of the plates, do they have to be absolutely the same? meaning that can one be smaller or larger than the other?
3) When finding the area of the plates, are the thicknesses of the plates taken into account. (I know Area is L * W, and adding the thickness would mean Volume; however, since voltage is flowing through the plates, i wonder about it.)
4) Since the units of length that i am obligated to use are in inches would that interfere with the equations that are given for Capacitance such as:

C = εrε0[itex]\frac{A}{d}[/itex]

C = εr[itex]\frac{A}{4*Pi*d}[/itex]


5) One final question, for the equation would be why is the 4*pi applied in the second equation or is it used to omit ε0 constant?

Please and thank you in advance for anyone's time.

Welcome to the PF.

The plates do not have to be identical, but you will get a capacitance closest to that first equation if the plates are the same area and are spaced close together compared to their 2-D size.

You do not take the thickness of the plates into account. The first equation is correct -- I'm not sure where the 2nd equation came from, but at least when using MKS units, the first equation is used.

And just because you are given the spacing of the plates as 0.062", that doesn't mean that you can't just convert that into meters and do the rest of the calculations in MKS units. After all, Farads are MKS units. :smile:
 

1. What is a parallel plate capacitor?

A parallel plate capacitor is a device used to store electrical energy. It consists of two parallel conducting plates separated by a dielectric material. When a voltage is applied to the plates, an electric field is created between them, causing the plates to store opposite charges.

2. How does the parallel plate capacitor model work?

The parallel plate capacitor model is based on the principle that the capacitance (C) of a capacitor is directly proportional to the surface area of the plates (A) and inversely proportional to the distance between them (d). This can be represented by the equation C = ε0A/d, where ε0 is the permittivity of free space.

3. What factors affect the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is affected by the surface area of the plates, the distance between them, and the type of dielectric material used. It is also influenced by the applied voltage and the frequency of the alternating current passing through the capacitor.

4. How is the capacitance of a parallel plate capacitor calculated?

The capacitance of a parallel plate capacitor can be calculated using the equation C = ε0A/d, where ε0 is the permittivity of free space, A is the surface area of the plates, and d is the distance between the plates. The unit of capacitance is Farads (F).

5. How is the parallel plate capacitor model used in practical applications?

The parallel plate capacitor model is used in various practical applications such as in electronic circuits, power supplies, and energy storage devices. It is also used in sensors, filters, and other electronic components. The model helps in understanding the behavior of capacitors and designing circuits for specific purposes.

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