Round Plate Capacitor?

[gladiator98]

How would I analyze a round plate capacitor (capacitance, energy stored etc.)? Would it be similar to the flat plate capacitor analysis? Thank you. Picture Attached ^

 

[MrAl]

Hi,

When you say "round" do you mean two half cylinders or two half spheres, or even some other shapes?

 

[gladiator98]

Two half cylinders; sorry about not making that clearer. I'm just guessing I can model these two as flat plate capacitors but since it's round it'll have different areas with the distance between the two as 2*Radius. I dont know though need your input. Ty.

 

[MrAl]

Hi again,


To solve this kind of problem i think you have to solve the right Laplace equation, but you'll have to look that up on the web. I can help with an approximation however which is probably going to be good enough for many cases.

If you look at the attached diagram, Cap 1 is a standard parallel plate capacitor with plates separated by a distance D and separator with dielectric constant er. You can see that the formula involves the distance D and the area A and for Cap 1 and both D and A are constants. Cap 2 on the other hand has a varied distance along its plate width where the distance on the left side is H1 and on the right side H2, which means its capacitance varies along the width (in the direction of x). If we divide the capacitor into 'N' smaller parts and number each part 'k' then each area we can call Ak and each distance Dk. We then develop formulas for Ak and Dk and sum all of the smaller parts, eventually letting N approach infinity and finding at last the infinite sum which would be the total capacitance. Of course the field will fringe more on the right than on the left, so this will be an approximation.

If this is a technique you would like to pursue, we can go farther with it. The attached diagram shows the start of working with a capacitor that has slanted plates rather than parallel. A *very* rough approximation to YOUR problem would be two of these side by side, with the second one flipped horizontally so that the upper part forms an upside down 'V' shape and the lower part a 'V' shape, and the symmetricalness is basically the same. The idea would be to form a better and better approximation starting with that one.

A question you could answer is just how 'curved' are the plates of your capacitor...do they actually form two halves of a cylinder or are they two halves made from only a small slice of a cylinder such that they are only curved a little bit.

Added later:
Instead of letting N go to infinity, we'll let dx go to zero instead after defining everything in terms of dx.

 

[gladiator98]

Thanks for your response! The plates are near perfect circle cylinder halves. I'm still not sure where to go lol.

 

[MrAl]

Hello again,


Let me illustrate the slanted plate capacitor and you can take it from there. Note that the dimensions are outlined in my previous post attachement, the capacitor labeled "Cap 2".

We start with the parallel plate capacitor formula, which is:
Cp=Ke0*A/D
where
Cp is the capacitance,
Ke0 is er*e0,
A is area,
D is distance between plates,
er is dielectric constant,
e0 is free space permittivity.


We want to break the slanted cap down into smaller sections and calculate each section separately, then add the results. If we break them down into infinitesimally small sections, we get an exact expression for the capacitance in the end although we have to remember that these techniques are only approximations to begin with just like the parallel plate capacitor.

Each small cap will have capacitance:
Ck=Ke0*Ak/Dk

where
Ak is the incremental area, and
Dk is the incremental distance which changes over the width x.

We'll declare the increment in x (usually called 'delta x') as 'dx', and define everything in terms of dx or x alone. When we let dx approach zero we get an exact expression because as dx goes to zero the error in the calculation of one Ck goes to zero.

The actual width is slanted and so has a slope M, and is related to dx by:
Wk=sqrt(dx^2+(M*dx)^2)
where M=H/W (H is the height of the plate alone and W is the width),
and this is simply the Pythagorean theorem applied to the base and rise, and simplifying we get:
Wk=dx*sqrt(M^2+1)
and all this means is that for any increment (dx) in the left to right distance (x) we have a plate width Wk.
To find the area Ak now all we have to do is multiply Wk times the length of the plate (the length of the plate is the distance across the plate looking into the page) and we get:
Ak=Wk*L
where
L is the length.

Now that we know what the plate width is relative to dx, we then write an expression for Dk the distance as that changes with x:
Dk=2*M*x+D1

where D1 is the same as H1, and H1 is the left hand side plate separation, and note this is just the equation for a line y=m*x+b applied to the two slants.

Now recalling the incremental capacitance formula:
Ck=Ke0*Ak/Dk
and substituting Ak and Dk with the expressions we formed above, we collectively end up with:
Ck=dx*Ke0*L*sqrt(M^2+1)/(2*x*M+D1)

Note that at this point if we call dx "delta x" and make dx small (but not zero) and sum all the Ck we would get an approximation:
Sum (Ck)=Sum (dx*Ke0*L*sqrt(M^2+1)/(2*x*M+D1))
so if we let dx=W/10 for example (break the cap up into 10 smaller caps) we would get a rough approximation although there would be a small error in the calculation of the total C.

But taking the limit as dx approaches zero we get an exact expression:
C=integral[0 to W](Ke0*L*sqrt(M^2+1)/(2*x*M+D1)) dx
where W is the total width of the capacitor plates.

After computing that integral, whatever that might take, we end up with a nice little expression:
C=(Ke0*L*sqrt(M^2+1)*(ln(2*M*W+D1)-ln(D1)))/(2*M)

As a basic sanity check, if we make the plate height H zero we end up with a parallel plate capacitor again, so if we take the limit of the above expression as the height H approaches zero we should get the same expression back again as for the parallel plate capacitor:

Cp=lim[H-->0]{ (Ke0*L*sqrt(M^2+1)*(ln(2*M*W+D1)-ln(D1)))/(2*M)}

and this is what we get:
Cp=(Ke0*L*W)/D1

which is indeed the formula for the parallel plate capacitor with A=L*W and D=D1.

 

[gladiator98]

Well, I modeled it closely to a cylindrical capacitor like a coaxial cable. Instead my answer came out a little differently of course. I got:

C = (PI*ε*L)/(Ln(r))

compared to

C = (2*PI*ε*L)/(ln(b/a)) for a coaxial cable

Reference: **broken link removed**

 

[gladiator98]

But, it's undefined at r = 1