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Asoom · May 9, 2009, 10:55 AM

a) The area of parallel -plate capacitor is 5.0 cm^2 and is filled completely with Teflon. If the separtion between the plates is 2.6mm, what is the maximum energy that can be stored before break down ?

b) The capcitor is now charged to half Qmax and then disconnected from the battery. What is the energy stored in the capacitor and the voltage across its plates ?

I solve first one and I get 0.2165 J

but the second one I can't solve it , so anyone can help me ?

bhysics · May 10, 2009, 10:09 AM

This is a basic problem...
You have just to combine the two important formulae from the capacitors...

Perito · May 10, 2009, 10:40 AM

Capacitance is given by:

$C = \frac QV$

or

$C = \epsilon_r \frac Ad$

The energy stored is

$E = \frac 12 CV^2$

C is the capacitance in farads, Q is the charge in coulombs, V is the voltage in volts, A is the area, d is the distance between the plates, εr is the dielectric constant (static relative permittivity)

That should give you almost everything you need to solve the problems. You'll need the dielectric constant for Teflon. Google "relative static permeability" and you'll find it.

Here is a site with a lot of calculators and explanations about capacitors. You should be able to find the calculator you're looking for. Plug your values in and it'll tell you if you calculated correctly.

http://hyperphysics.phy-astr.gsu.edu...capcon.html#c1

bhysics · May 10, 2009, 12:28 PM

like I said^^