Engineering Calculating the gap of a parallel plate capacitor

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SUMMARY

The discussion centers on calculating the gap of a parallel plate capacitor, specifically using the dielectric constant of 2.55 and circular plates with a radius of 28 mm. Participants clarify that the capacitance formula can be rearranged to solve for the gap distance (d), emphasizing the importance of the ratio of the gap to the plate diameter for accuracy. A calculation example is provided, demonstrating that the initial result of 0.46 mm must be multiplied by 10 to yield the correct gap of 4.6 mm.

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joemte
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Homework Statement
Calculate the gap required between the plates to give the component a capacitance of 12pF to one decimal place. (mm)
Relevant Equations
permittivity of a vacuum (𝜀0) to be 8.885×10−12𝐹 𝑚−1, 𝜋 to be 3.142
I've been given this question for my TMA2, I've tried looking at the learning material but it gives no information on how to calculate the gap? Does anyone have a formula for this? Or can someone point me in the right direction?

Thanks
 
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Welcome to the PF. :smile:

Hint -- do a search in the topic of capacitors at Wikipedia, and you will find the equation you are looking for about half-way down the article (near the diagram below).

Also, the problem as stated does not have enough information for you to calculate the gap distance. Given a gap distance, the capacitance goes up for larger plate areas. Are you given the plate area in the problem statement?

1568729986274.png
 
Hi! Thanks,

Also, yes sorry - Dielectric constant of 2.55. The capacitor is constructed with circular plates of radius 28 mm.

I saw equations for voltage, magnitude of electric field, capacitance and maximum energy, but not one for calculating the gap required. Or would I need to transpose this formula (Capacitance) for d?

{\displaystyle C={\varepsilon A \over d}}
 
The variable "d" in that equation is the gap, yes. Does that make sense? And the gap is filled with the dielectric material like in the diagram above, right? :smile:

There is actually a subtlety about the capacitance value of a parallel plate capacitor -- the simple equation above applies when the gap is much, much less than the plate dimensions. So if the gap is less than 1% of the diameter of your circular parallel plate capacitor, the number should be accurate to a percent or so. It's interesting that your problem statement asked for a particular accuracy.

So when you do the calculations, also take the ratio of the gap d divided by the plate diameter to see if the approximate equation above is valid for your problem. If it is not, we can help you find a much more complicated calculation to do...
 
Hi,

Thanks for your help it finally clicked, but I'm not sure where i am going wrong to have to times my answer by 10 to make it correct.

(2.55)(8.885*10^-12)Pi 0.28^2)
___________________________________ This gives me 0.46, I multiply the answer by 10 and it is correct (4.6mm)
1.2*10^-11
 
joemte said:
circular plates of radius 28 mm
joemte said:
Pi 0.28^2
28mm is how many meters? :smile:
 

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