What is the maximum energy that can be stored by the capacitor?

Question

The dielectric in a capacitor serves two purposes. It increases the capacitance, compared to an otherwise identical capacitor with an air gap, and it increases the maximum potential difference the capacitor can support. Recall that air near the sphere of a Van de Graaff generator undergoes breakdown, creating a spark, when the electric field is sufficiently strong. The same is true for any material. That critical field strength, at which breakdown occurs, is 3.0 MV/m for air, but 60 MV/m for Teflon.

a) A parallel plate capacitor consists of two 4.0cm X 4.0cm plates spaced 0.70mm apart with only air between them. What is the maximum energy that can be stored by the capacitor?

b) What is the maximum energy that can be stored if the plates are separated by a 0.70mm-thick Teflon sheet?

Answer 1

Ginger...

(1)recall that the energy in Joules stored in a capacitor is the capacitance (C) times the voltage voltage squared (V^2) or CV^2

(2)  Capacitance is calculated by

C= K*Eo*A/D, where  Eo= 8.854x10-12

(using millimeters for area  distance)

where:

K is the dielectric constant of the material,

A is the overlapping surface area of the plates,

d is the distance between the plates, and

C is capacitance

(3) If you plugin the numbers from a in your question...and use a dielectric constant of air as 1...then the Capacitance is 20.2 pf (picofarads)

(4) Now you know the distance is .70mm so you can figure the max breakdown voltage for air because they gave you dielectric breakdown voltage of 3.0Mv/m (million volts per meter)...so what's the breakdown voltage at .70mm

.70mm X 1m/1000mm = .0007m (distance in meters)

3.0MV/m X .0007m = 2100volts

so the air's max voltage at .70mm is 2100volts...past this you will get an arc (break down of dielectric)

(5) Now that you know the breakdown of air...2100volts...and the capacitance 20.2 pf...you can calculate the max energy that can be stored in this air dielectric capacitor

Energy = CV^2 = 20.2 pf X 2100^2=.089082 Joules

OK...that's the answer to a...so you can do 'b'...hint dielectric constant of teflon is about twice as much as air so your answer should be about twice as much...(use the dielectric constants the class provided you with or look up....or use mine that I retrieved from my 30 year memory)

Good Luck!

Answer 2

Cute. So what is your problem?

First you have to figure out the capacitance of the air capacitor, call it Ca.

http://hyperphysics.phy-astr.gsu.edu/hba...

Then, given the plate distance and the critical field strength, you can figure out the maximum voltage you can apply. (i.e. for a distance of D meters and a critical field strength of V per meter, the voltage you can use is Va = DV)

With Ca and Va you can figure the energy stored for the air cap.

http://en.wikipedia.org/wiki/Capacitance

Putting teflon between the plates changes the capacitance by the factor of the dielectric constant

http://physics.bu.edu/~duffy/semester2/c...

and changes the critical field strength, without changing the distance between the plates, so you can compute the capacitance with teflon, Ct and the maximum usable voltage Vt.

With these you can compute the energy that can be stored.