Multiple dielectrics in one capacitor

[Rake-MC]

Hi guys, I'm absolutely, completely and utterly stumped with this one question I've been given..

Here is the exact question:

Given that you possesses the following materials:
(i) two sheets of copper of uniform thickness;
(ii) one sheet of mica of uniform thickness 0.10 mm and dielectric constant κ= 6;
(iii) one sheet of glass of uniform thickness 2.0 mm and dielectric constant κ = 7;
(iv) one sheet of paraffin wax of uniform thickness 1.0 cm and dielectric constant
κ = 2;
determine which sheet, or combination of sheets, will produce a parallel plate capacitor of
the largest capacitance when placed between the copper sheets. Assume that all sheets
have the same shape.

Relevant equations that I have are:
C=(kappa)(epsilon)(Area)/(Distance)
V=V[initial]/kappa
but there are probably plenty more. I have equations for potential energy and such, but I'm not sure if they are necessary for the question.

attempt at a solution

Well as I said, I am completely stumped. I can calculate the capacitance of each dielectric on its own in the capacitor, but the combination ones are the ones I'm stuck with. This is what I've got so far {probably way off track}.

C[mica] = 6(epsilon)(area)/0.1x10^-3
C[glass] = 7(epsilon)(area)/2x10^-3
C[wax] = 2(epsilon)(area)/0.01

Can I ignore epsilon and area because they are common in all the equations?

for the dielectrics next to each other;

C[effective] = C[no dielectric]*k[1]*k[2]

This is most likely wrong. I know that for one dielectric, you can simply say that C[effective] = C[no dielectric]*k
where k = kappa

But I simply don't know how to derive a formula for multiple dielectrics.

Also, I do have knowledge of the theory of how dielectrics work, ie. how they generate opposing electric fields to that of the capacitor, but I am still 100% stuck with this question..Thanks in advance

 

[Irid]

If you have multiple sheets of dielectric stacked on each other, you can imagine that each sheet is a distinct capacitor and these capacitors are connected in series. The total capacitance C is then

$$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n}$$