1
$\begingroup$

I have this doubt wandering through my mind for a long time.

A capacitor, according to the definition of physics, is a charge storage device. In a simple parallel plate capacitor, when electrons accumulate on one side of the plate, they push away the electrons on the other side

Let's say there are 30 electrons on either sides. If voltage is applied, 30 electrons which accumulate on one side (say plate A) must push 30 electrons on the other side (plate B). Then current flows.

But if those 30 electrons travel through the circuit and reaches the plate A, to my knowledge, the 60 electrons stay in plate A. Without any electrons in circuit (since all electrons already accumulated in plate A), how will the current flow through the circuit.

I don't know whether this question is correct. Correct me if I am wrong. Hope you understand my question.

$\endgroup$
1
  • 5
    $\begingroup$ Consider displacement current while you are talking about "electron pushing". Anyways you are right about current will stop flowing. That happens when the capacitor is fully charged. $\endgroup$ Oct 22 '15 at 13:33
1
$\begingroup$

I cannot completely follow your approach to the capacitor. However, I want to explain mine to you:

Two parallel plates, acting as a condensator that is not charged, will have your equal amount of electons on both sides. There is no energy stored in this state.

Now you need to add work to the system to separate the electrons on one side from their atoms and bring them to the other side, hence charging the plates and creating an electric field between them. The accumulated electrons actually don't want to be close to each other, because they are of the same charge. They try to get as close as possible to the inverse charge at the other plate front, which is why they accumulate at their own plate's front. This state stores the energy as long as the plates are not connected, or an outside energy source is adding potential to keep the state alive.

Now at this point you can release the energy by connecting the two plates (as the electrons now see a way to get to their opposite charges) and use it to power a circuit with the potential energy which you have added in the previous step.

So if I understand your question correctly, the misunderstanding is placed at when you need to put work to the condensator and the point when you can use the energy for a circuit.

$\endgroup$
2
  • $\begingroup$ What kind of work? Is it electric current? $\endgroup$ Oct 24 '15 at 9:08
  • $\begingroup$ Work is work, and current is current. You can transform different kinds of energy into another one. In this case the potential energy. The energy you transform is given by P=U*I, and will be dissipated as heat, mainly due to electrical resistivity as in ohm's law R = U/I, or, if you include motors or comparable active devices, as kinetic, potential, electromagnetic energy. $\endgroup$
    – kamuro
    Oct 26 '15 at 10:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.