Basically, a simple capacitor will consist of 2 plates and a dielectric material between them, what if i took 2 plates, charged one with a +ve charge and the other with equal -ve charge, then i put them together with a dielectric material between them , then i connect them with a wire so it can discharge , can i consider this system a capacitor and thus using E=1/2 *C *V^2

  • $\begingroup$ Note, that this is a duplicate of physics.stackexchange.com/q/101116. $\endgroup$ – Tobias Mar 1 '14 at 6:15
  • $\begingroup$ How is this question a duplicate of thel inked ? @Tobias $\endgroup$ – Buraian Apr 22 at 20:03

The plates form a capacitor from the beginning. (More precisely a coupled capacitor with three terminals because you have also to regard the neutral ground as one of the terminals.)

The capacity coefficient between the plates is low at the beginning since the plates have a large distance. Therefore, you need a high voltage to put the charge on them.

In the process of bringing the plates close together you actually gain mechanical work from the system since the plates do attract each other and you move the plates in the direction of the attracting electrostatic force.

The done mechanical work reduces the energy of the system. This is reflected by a decrease of capacitor voltage.

In electrical terms the voltage drops because the charge keeps constant and the capacity grows.

Answer to the question from the comment:

Charging is actually separation of positive and negative charges. That means you cannot separately charge one plate with a negative charge. The corresponding positive charge must be somewhere.

For an instance you can put the corresponding positive charge on the ground. If you insulate the first charged plate from the ground you can switch off the energy supply and the first insulated plate will keep its charge.

If the second plate has been insulated before the experiment then it will not be charged.

If you bring the second plate close to the first one there will be some positive surface charge on the surface at the side directed towards the first plate but this surface charge will be compensated by a negative surface charge on the other side of the second plate. So the net charge on the second plate will be zero.

If you now ground the second plate (which is still very close to the insulated first one) then some electrons will flow from the second plate to the ground to compensate the positive ground charge. (You can imagine that as positive charge flowing from the ground towards the second plate.) This way the system will assume the state of lowest electrical energy. (A part of the energy will be transformed into heat caused by the current and the electrical resistance of the ground.)

In physics one considers a single point charge. This is an abstraction. One only considers the field in the proximity of a very small object that is charged w.r.t. the ground. The distance of the observation point from the ground is much larger than the distance from the small object. Therefore, the actual shape of the ground does not influence the field at the observation point very much and the single point charge model gives a good approximation of the field.

  • $\begingroup$ thanks, i have another question though , can i charge a capacitor from one plate only? for example: if i charge one plate with-ve charges and i ground the other plate, it will gain +ve after unplugging the -ve charge source and the ground , will it act like a normal capacitor? $\endgroup$ – user28324 Feb 28 '14 at 21:53
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    $\begingroup$ @user28324 Yes you charge a capacitor from one plate only. By putting negative charge on one plate the other plate will automatically place positive charge. $\endgroup$ – jerk_dadt Feb 28 '14 at 22:33
  • $\begingroup$ @user28324 I have extended my answer. Now it also addresses you second question from the comment. $\endgroup$ – Tobias Feb 28 '14 at 22:53

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