# Electrical Engineering - Design Problem - Capacitors

1. Mar 12, 2013

### GreenPrint

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!!!

2. Mar 16, 2013

### GreenPrint

Can I post this in the introductory physics forum?