Energy in a charged metal plate

1. Oct 29, 2008

raybuzz

Hi all,
If i have a charged plate of charge Q coulumbs, then what is the energy contained in it?
Is the energy independent of the way in which i managed to get a single plate of Q columbs charge? In the sense , suppose in the beginning there were two plates of charge +Q, -Q columbs and distance r apart, then i manage to apply mechanical energy against the electrostatic force of attraction between the plates and i seperate it to large( say infinite) distance apart, then the mechanical energy spent will be--
Q^2/ 4pi(epsilon o)r

from law of conservation of energy this mechanical energy spent in seperating the plates must be stored as electric energy?? I am confused please help.

2. Oct 29, 2008

Naty1

The energy stored in a capacitor is 1/2CV^2....

for a parallel plate configuration, C= keA/d.

So yes, this is the potential energy, the amount of work, in theory required to separate the plates to infinity....to pull the charges apart....or it's the amount of energy recovered if the plates are allowed to come together...and it's the amount of chemical energy depleted from a battery charging the plates.

your stated formula looks like one for an isolated spherical configuration, hence "R":

The potential for an isolated conducting sphere is V = q/4pi eR, so W= qV becomes

W=q^2/4pi eR....but that's not for a parallel plate capacitor...

3. Oct 30, 2008

raybuzz

Ok, say the energy is .5CV^2
Then the energy required to seperate it is .5CV^2.
From the law of conservation of energy this energy is converted to another form.
I want to know if it is infact stored as electrical energy in the two charged plates.
There doesnt seem to be a way in which we can find the energy on a charged, single plate.