Okay, so I know that a capacitor is charged by the battery providing a potential difference which forces positive charge to flow from one capacitor plate to the other (Conventional Flow). This flow of charges takes place until the steady state is achieved.


1)Why do we calculate energy stored in the capacitor by assuming that a small charge $\text dq$ is displaced from one plate to the other ?

2)Does the charge actually flow through air (No dielectric)

3) Let us say that a charge of $+q$ and $-q$ has appeared on the capacitor plates and the steady state is not yet achieved. We say that a charge $\text dq$ will flow from the low potential plate to the negative terminal of the battery. The battery increase its potential by $V*\text dq$ ($V$ is the EMF) and sends it to the high potential plate. Is it absolutely necessary that THIS charge travel from the the negative plate to the positive one. I have been taught that a wire has a large number of free charge carriers (positive charge in case of conventional flow). Can t the $\text dq$ charge originate from the wire itself . If this is true then then shouldn’t the work required to move this charge to the the positive plate change since now it has NOT traveled through a potential difference of $q/C$ ($C$ is the capacitance of the capacitor).


You are correct in saying that a charge on one plate is probably not brought all the way to the other plate (or at least there is no way to confirm this I suppose (or at least I can't think of a reason why you would want to or need to confirm this)). However, if you want to say, "ok, well really this charge $q_1$ that started at this plate will move just a little closer to the other plate," then you will also say, "ok, well this charge $q_2$ that started a little closer to the other plate compared to that charge $q_1$ that was on the plate will move even closer to the other plate," and so on and so on. So really, you can either break it all up charge by charge, or you can just think of it as moving a single charge (or set of charges). The energy to do this will end up being the same either way. It's easier to just think of moving charges from one plate to another, so this is usually how it is discussed.

Typically the charges do not flow through the air, they flow through the circuit. But since the electrostatic force is conservative, we don't really care how a charge got to where it is from where it started. It is sufficient just to know where it started and where it ended to know its change in potential energy. We can choose any path we want to. So we pick one that just goes directly from one plate to the other plate, as this analysis is fairly simple for, say, a parallel plate capacitor.

  • $\begingroup$ okay so while calculating the energy of a capacitor we consider that a small charge having the same potential as the low potential plate is taken to a higher potential . We consider this potential difference to be the difference in potentials of the capacitor plates. But if the same charge is not taken from the low potential plate to the high potential plate, wont the potential difference change since we don’t know anything about the initial potential of the charge ? $\endgroup$ – Aditya Ahuja Nov 1 '19 at 4:04
  • $\begingroup$ @AdityaAhuja I'm not sure I'm following you here $\endgroup$ – BioPhysicist Nov 1 '19 at 4:30
  • $\begingroup$ Consider the plate at lower potential to be A and the one at higher potential at B. When we move a charge from A to B we say that a work has been done against a potential difference of V (low potential to high potential ) this charge is q1 according to your answer. What confuses me is the charge q2 which is closer to the high potential plate . How can we say that this charge q2 has also moved against a potential difference of V since it has not originated from the low potential plate? $\endgroup$ – Aditya Ahuja Nov 1 '19 at 5:05
  • $\begingroup$ @AdityaAhuja We usually approximate wires in a circuit as being equipotential surfaces. The interesting things with changes in potential happen at resistors, batteries, etc. So if you want to think about just a battery and capacitor, really all of the work is just done at the battery. But the idea is the same. As charges go across the battery they gain energy. You can either think of moving one charge the entire way or moving many charges a little bit. $\endgroup$ – BioPhysicist Nov 1 '19 at 5:34
  • $\begingroup$ I think I get it now. So as the potential of the low potential plate changes the potential of the wire it is connected to also changes ? $\endgroup$ – Aditya Ahuja Nov 1 '19 at 10:58

Is it absolutely necessary that THIS charge travel from the the negative plate to the positive one.


To reinforce @AAron Stevens answer, it is even possible that no specific charge (electron) will travel from the negative plate to the positive one when charging a capacitor because the drift velocity of the charges is very low, depending on the current and the size, type and length of the conductors connecting the battery to the capacitor.

For example, let's say the initial charging current of the capacitor is 1 ampere and that the battery is connected to the capacitor plates by 2 mm dia copper conductors. The average drift velocity of the electrons in the copper conductors will be about 2.3 x 10$^{-5}$ m/s. Which means on average a single electron would take about 16 minutes to travel a distance of 2.3 mm in the copper conductor.

Of course as Aaron pointed out it is not necessary for any specific charge to go from one plate to the other. While the drift velocity of the charges is low, all the charges in the conductor experience the electric field almost immediately (nearly the speed of light). As soon as the field is applied, all of the charges start moving in a chain reaction, resulting in charges near/on plate being delivered/removed as soon as the field is applied.


In response to your comment that my answer didn’t provide an energy analysis, it was because I felt @Aaron Stevens already covered it. But let me give you a rough analogy with gravity.

Let’s say I have 1000, 1kg rocks in a 1 m tall cylinder. I raise the cylinder 1 mm in 1 second. The total increase in gravitational potential energy of the collection of rocks is mgh or 1000 x 9.81 x 0.001 = 9.81 kg m$^2$/s$^2$ or 9.81 N.m. Although no single 1 kg rock traveled 1 meter, the increase in gravitational potential energy of the collection of rocks is equivalent to a single 1 kg rock being raised 1 meter. The gravitational potential difference (work per unit mass) is 9.81 m$^2$/s$^2$

The electrical analogy: Each 1 kg rock is analogous to 1 Coulomb of charge. 9.8 m$^2$/$s^2$ is analogous to voltage (work done per unit charge to move the charge 1 meter, or 1 volt) and raising the cylinder 1 mm in one second is analogous to current (1 ampere).

Hope this helps.

  • $\begingroup$ This answer does not have an analysis of the energy of the capacitor. $\endgroup$ – Aditya Ahuja Nov 1 '19 at 5:07
  • $\begingroup$ @Aditya Ahuja I felt Aaron Stevens already covered it. But I will provide analogy with gravity $\endgroup$ – Bob D Nov 1 '19 at 7:16
  • $\begingroup$ @Aditya Ahuja See my updated answer $\endgroup$ – Bob D Nov 1 '19 at 8:22
  • $\begingroup$ But how is this analogy applicable to an RC circuit ? It can easily be seen that each rock is raised by 1mm. But how can we say that the EACH charge (rocks according to your analogy) in case of a circuit moves against a potential difference of V (potential between the plates). Isn’t this right only when a charge on the negative plate (low potential plate) is taken all the way to the positive plate (high potential plate) ? $\endgroup$ – Aditya Ahuja Nov 1 '19 at 15:18
  • $\begingroup$ @AdityaAhuja But can’t you see that it is not just one rock that has to be raised 1 mm, but all 1000 rocks must be raised 1 mm in order to deposit the top rock (one coulomb of charge) on the + plate (plate at high potential) and simultaneously remove the bottom rock (1 coulomb of charge) from the – plate (plate at low potential)? This is dictated by the extremely slow average drift velocity of the charges in the conductors. $\endgroup$ – Bob D Nov 1 '19 at 18:12

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