# Homework Help: Find thickness of polystyrene

1. Dec 18, 2012

### warnexus

1. The problem statement, all variables and given/known data

2. Relevant equations

Capacitance = k(epsilon zero) (Area) / (distance)

Length (thickness) = (Volt)(Area) / (Current * Resistivity)

3. The attempt at a solution

Given:

Capacitance : 480 * 10 ^ -12 Farad
Radius: 24 * 10 ^ - 2 m
Area: pi * r ^ 2
dielectric constant of polystyrene: 2.6
breakdown field(MV/m) : 25
Resistivity of Polystyrene : 10 ^ 15

distance = (2.6)(8.85 * 10 ^ -12) (3.14 * (24* 10 ^ -2) ^ 2) / (480 * 10 ^ -12)

however finding the distance does not seem to help me at all since the other equation do not involve distance

any guidance would help

Last edited: Dec 18, 2012
2. Dec 18, 2012

### TSny

Hello. What distance does this represent physically?

3. Dec 18, 2012

### warnexus

distance apart for two capacitors

4. Dec 18, 2012

### TSny

I guess you mean the distance between the two circular plates of the capacitor. (You only have one capacitor.)

But isn't this the same as the thickness of the polystyrene?

5. Dec 18, 2012

### warnexus

oh so thats what they mean! now that I think about you have a point there. so length, spacing, distance,thickness are same ideas. thanks for the guidance

Last edited: Dec 18, 2012
6. Dec 18, 2012

### TSny

Yes. It's always important to know what the symbols in an equation represent.

7. Dec 18, 2012

### warnexus

I was wondering. I need to find voltage of the capacitor. I do not know the current.

resistance = (resistivity)(length)/(pi * (r)^ 2)

I wind up finding the resistance which turned out to be 4793.7 ohms or 4793 ohms

but current is still missing.

current = (voltage)/(resistance)

voltage = (dielectric constant)(epsilon zero) (Current)(Resistivity)/(Capacitance)

or

I could have found voltage a different way

voltage = (Charge)/(Capacitance)

but no information about what charge is within the capacitor

Last edited: Dec 18, 2012
8. Dec 18, 2012

### haruspex

It's a capacitor; you do not want current.
What do you take the term 'breakdown voltage' to mean?

9. Dec 18, 2012

### warnexus

failure of an electrical circuit

well i did list the polystyrene breakdown field to be 25 * 10 ^ 6 V/m

oh now I get it! multiply the thickness by the breakdown voltage and viola working voltage. thanks haruspex!

Last edited: Dec 18, 2012
10. Dec 18, 2012

### haruspex

But what specifically in relation to capacitors?
Yes, but you did not use it.
So what would the breakdown voltage be for this capacitor?

11. Dec 18, 2012

### warnexus

failure of an electrical field from a capacitor

i modify the content of my previous post. the electric field equation Electrical Field = (Volt/Area) would not work. I just wind up using the answer I got for thickness and used it with the breakdown voltage and found voltage that way.

12. Dec 19, 2012

### haruspex

No, it would be the capacitor passing a DC current, which it is not supposed to do.
Let's get the terminology right: you used the breakdown field and the polystyrene thickness to find the breakdown voltage, which is the 'working voltage' required, yes?