What is the Capacitance of a Multi Plate Capacitor?

[Callum10]

Hi,

The problem is as follows:

There's a multi plate capacitor, with plates of side L, equidistant from each other, but the plates get increasingly longer as a function of horizontal distance from the first plate, starting at height A, and finishing at height B.

In other words, the first plate at x=0 has height A, and the last plate at x=B-A has height B. There are a large number of plates in this configuration. We are asked to show the capacitance of this is (ε0 L) / ln(B/A) where ε0 is epsilon naught. Has anyone got any ideas how to even start this off?

I've attached the question because my explanation of it is really bad!

Any help is much appreciated,

thanks

 

[Isterix]

Here's thought: The first capacitor is of length A (you neglect the extra height of the second plate), the second capacitor is the length of the second plate and soo own.
You calculate the general function for a capacitor with length d (a<d<b) and then integrate!
I think... :)

 

[Callum10]

but what's the length of the second plate?

 

[Isterix]

the length of a plate at distance d from the first plate is c1*d (where c1 is a constant)
and you know that c=(ε0*S)/l
s= surface area of the plates
l=the distence between two plates
and you can consider all the capacitors are connected in sequece (one after another - I really don't know the terminology as well as I should)

 

[tutor69]

Since all the capacitors can be considered to be in series the equation to find the C
$$C \sub{tot} = (C1)^{-1}+(C2)^{-1}+...+(Cn)^{-1}$$

Next use C=(ε0*S)/d
where d is the distance between two plates.

Given that height of plate is a linear function of x, like h = a + x
as at x=0, h=a and at x= b-a, h=b

Apply these two facts under the limit d---> 0 and n, number of plates ---> infinity.The summation in first equation will approach to an integral from x = 0 to x = b-a

Try to find out total C

 

[Callum10]

I understand everything that you've written down, but don't see how it all fits together to get the final total capacitance

thanks for the help

 

[Isterix]

as far as I know (ctot)^-1= c1^-1+...

 

[tutor69]

yes you are right . I made a mistake ther i am editing it :

$$(C \sub{tot})^{-1} = (C1)^{-1}+(C2)^{-1}+...+(Cn)^{-1}$$

Callum 10, you have to do a little work with the information. S = Area = l * h
where l is the width of the plates given , equal for all plates. h = a + x
Remember that first plate is at x=0 and with interplate distance of d each it reaches to x= b-a for the final plates. Now the problem says that this interplate distance is very small there fore we can assume d or dx ( since this is an incremental value in x also) approaching zero.

How do you define an integral? it is limit of a sum
here is how you can convert above sum i.e. Ctot into an integral which will yield you the desired result. I hope now I clarified you the steps. and that you can solve the integral.