Electrical Engineering - Design Problem - Capacitors

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Problem
Given the power that is needed to be output by the capacitor - P
the length of time that the capictor must give off that power - t
the maximum operation voltage that the capacitor can operate at- V
the dialectric constant of several materials - ε
the breakdown field of those several materials - $U_{d}$
and the mass density of those several materials - ρ
Design the capacitor that meets these requirements

Dialectic Material 1
ε= $4ε_{0}$
$U_{d} = 1 \frac{V}{nm}$
$ρ = 2.5 \frac{g}{cm^{3}}$

Dialectic Material 2
ε= $80ε_{0}$
$U_{d} = .03 \frac{V}{nm}$
$ρ = 1 \frac{g}{cm^{3}}$

Dialectic Material 3
ε= $200ε_{0}$
$U_{d} = .1 \frac{V}{nm}$
$ρ = 1 \frac{g}{cm^{3}}$

Metal 1
$ρ=2.7 \frac{g}{cm^{3}}$

Metal 2
$ρ=2.1 \frac{g}{cm^{3}}$

The machine which can make the capacitor has the following thickness restrictions
$10 nm < t_{dialectic}< 10 μm$
$10 μm < t_{dialectic}< 10 cm$

Equations
$C = \frac{εA}{d}$
$C = \frac{dQ}{dV}$
$U = \frac{dW}{dt}$
the maximum energy stored in a capacitor is below
$w = \frac{1}{2}εCd^{2}U_{d}^{2}$
$V = U_{d}d$
$U=CV(t)\frac{dV}{dt}$
$U=\frac{1}{2}CV^{2}$
$I=\frac{dQ}{dt}$
$hp ≈ 746 W$
$ε_{0} ≈ 8.85*10^{-12} \frac{F}{m}$
$m = 10^{9} nm$
$m = 10^{6} μm$
$m = 10^{3} cm$
$kg = 1000 g$

Attempt at a solution
I have no idea how to do this at all or where to begin. I just randomly chose material 3 for the dielectric material and started solving and go the following relations.

C = 3.082644628 F
Q = 678.1818182 C
$w = 10 ε_{0}Cd^{2}$

I don't know where to go from here or if I'm even on the right track at all. Thanks for any help that anyone can provide. It would be greatly appreciated.

Thanks!

 

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Can I post this in the introductory physics forum?