Energy in a charged metal plate

[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 separate 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.

 

[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...

 

[raybuzz]

Ok, say the energy is .5CV^2
Then the energy required to separate 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 doesn't seem to be a way in which we can find the energy on a charged, single plate.