Dielectric Breakdown-max potential difference

[PennStateFan1]

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

 

[gneill]

Your "first attempt" looks like the better way to proceed; It finds Vmax for the given plate separation from the dielectric breakdown field strength.

 

[PennStateFan1]

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?

 

[gneill]

I dunno. 3.86 x 10⁴ m² looks good to me.

 

[rwisz]

!
Dielectric breakdown is a phenomenon that occurs in insulating materials, such as air, when the electric field strength becomes too high and causes the material to lose its insulating properties. In the case of a parallel plate capacitor, this can happen when the potential difference between the plates reaches a certain maximum value, known as the dielectric breakdown voltage.

In part A, you correctly used the formula E=1/2CV^2 to find the required area of each plate, assuming a potential difference of 1.5 V. However, in part B, you need to consider the maximum potential difference that can be applied without causing dielectric breakdown. This means that the potential difference, Vmax, should be equal to the dielectric breakdown voltage, rather than 1.5 V. So your equation should be Vmax=E*d, where E is the maximum electric field strength for air (3E6 V/m) and d is the plate separation distance (2.21 mm). Solving for Vmax, we get a value of 6.63 V.

Now, using this value for Vmax, we can plug it into the equation E=1/2CV^2 and solve for C, the capacitance. Once we have the capacitance, we can then use it to find the required area of each plate using the formula C=εA/d, where ε is the permittivity of air (8.85E-12 F/m) and d is the plate separation distance. This will give us the area of each plate required to store the given energy of 3400 J with a maximum potential difference of 6.63 V.

I hope this helps clarify the problem and guide you towards the correct solution. Keep up the good work in your studies!