Parrallel Plate Capacitor Model Question

AI Thread Summary
The discussion centers on building a parallel plate capacitor with a target capacitance of 2000 microfarads and a plate separation of 0.062 inches. It clarifies that the plates do not need to be identical in material or dimensions, although similar sizes yield better results. The thickness of the plates is not considered in calculating capacitance, which relies on the area of the plates and their separation. Users are encouraged to convert measurements to MKS units for accurate calculations, as capacitance equations are typically expressed in these units. The second equation mentioned is not commonly used, and the first equation is preferred for MKS calculations.
UsualAnalyst
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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\frac{A}{d}

C = εr\frac{A}{4*Pi*d}


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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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\frac{A}{d}

C = εr\frac{A}{4*Pi*d}


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:
 
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