Dielectric Breakdown-max potential difference

AI Thread Summary
The discussion revolves around calculating the area of plates for a parallel plate capacitor designed to store energy equivalent to a typical AAA battery, with a focus on two scenarios: one at a potential difference of 1.5 V and another at the maximum voltage before dielectric breakdown occurs. The first part is solved using the energy formula, yielding an area of approximately 7.547E11 m². For the second part, the maximum voltage before breakdown is calculated to be 6630 V, leading to a required plate area of about 38613.1 m². Participants express confusion over the calculations and seek clarification on the methodology used. The thread highlights the importance of understanding dielectric breakdown limits in capacitor design.
PennStateFan1
Messages
8
Reaction score
0

Homework Statement



A typical AAA battery has stored energy of about 3400 J. (Battery capacity is typically listed as 625 mA h, meaning that much charge can be delivered at approximately 1.5 V.) Suppose you want to build a parallel plate capacitor to store this amount of energy, using a plate separation of 2.21 mm and with air filling the space between the plates.

a) Assuming that the potential difference across the capacitor is 1.5 V, what must the area of each plate be? I have the answer to this one 7.547E11 m^2

b) Assuming that the potential difference across the capacitor is the maximum that can be applied without dielectric breakdown occurring, what must the area of each plate be?

Homework Equations



For part A, I got the answer using the formula E=1/2CV^2

Part B I'm using Vmax=Emax*d
and E=1/2CV^2

The Attempt at a Solution


I've tried this two difference ways, as I'm kind of confused as to what its asking.

First attempt I'm using the value for Emax of 3E6 V/m for air and than multiplying that by .00221 m and then plugging it into the equation I used for part A

The other way is using Vmax is 1.5 Volts and solving for Emax and using that E as the E in the equation from part A.

Thanks for any help
 
Physics news on Phys.org
Your "first attempt" looks like the better way to proceed; It finds Vmax for the given plate separation from the dielectric breakdown field strength.
 
Using my first attempt

Vmax=6630 V

Then 3400=.5(C)(6630)^2
C=epsilon0*A/d=(8.854E-12)A/(.00221)=.000155
A=38613.1 m^2

Can you guys figure out what I'm doing wrong?
 
I dunno. 3.86 x 104 m2 looks good to me.
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top