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Work done by battery on a capacitor is QV/2, where V is the final potential across the capacitor plates an Q is the charge. I know that the Q charge which gets stored on the capacitor comes from the connecting wires. However, since
Positive charge on one plate is reducing (Assuming conventional flow) and increasing on the other it is convenient to assume that the charge is going directly from one plate to the other as it makes calculating work done on the charges more easier.

If we say some Q charge left one plate (let us call these charge carriers set A) and some charge Q ended up on the other plate (let this be B).Now, since it is not necessary that A went through the battery let us call the charge Q that went through the battery C. If battery didn't do work on A or B why do we say that the energy stored in the capacitor comes from the WORK done by the battery in transferring charges from one plate to the other.

According to my professor this work done by the battery can be assumed to be on the same charges since electric field is conservative and only depends on the initial and final states of the system.

I have assumed charges A,B and C to have the same magnitude. I know that distinguishing between charges is pointless but I have done so in order to make my question more clear.

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  • $\begingroup$ @ThePhoton $QV = CV^2$ $\endgroup$ – AgentS Dec 20 '19 at 19:46
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Energy is transferred from the battery to the capacitor. Work is defined as a transfer of energy. Therefore, the battery does work on the capacitor.

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  • $\begingroup$ But HOW is this energy transferred ? $\endgroup$ – Aditya Ahuja Dec 20 '19 at 19:06
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If the net charge on a plate is $0$, it takes no effort to move initial charge to that plate, so work done is $0$.

After you moved some charge to the plate, this excess charge on the plate opposes further new charge to be moved, so you must do positive work to overcome this opposition.

In a circuit, battery manages to move charge from one plate to the other. This means the battery is doing positive work against the electric field from existing net charge on the plate.

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  • $\begingroup$ How does this battery do work on charges which have have not gone through the battery. I have been taught that the battery only does work on charges crossing the battery. These charges then loose this energy while moving through the circuit. $\endgroup$ – Aditya Ahuja Dec 25 '19 at 21:00
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There are two ways to think about this.

  1. The work isn't done on the capacitor. It's done on the charge carriers that are pushed onto one capacitor plate and pulled off the other plate.

  2. The work is done to build the electric field between the capacitor plates, and energy is stored in the electric field.

Possibly the situation is more clear if you consider the 2nd version.

Even if the battery didn't do work on the exact carriers A that were pushed onto one of the capacitor plates, it did work on some carriers in the wire near its terminal, that did work on some other carriers along the wire, and so on until those carriers in group A got pushed on to the plate.

Similarly, when I push on the brake pedal in my car, I push some hydraulic fluid in a hose, which pushes other fluid along the brake lines, which makes the brake pads move against the disk or drum. Even though my foot didn't directly interact with the molecules of brake fluid that actually moved the pads, we can still say (speaking generally) that I did work on the pads to move them.

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  • $\begingroup$ taking some charge from battery's negative terminal to Positive terminal requires energy. This energy come from chemical reactions in the battery. The energy lost by the battery is given to the charge which went across the battery. Now let us consider some charge getting removed from the negative capacitor plate. How does this charge use up some of the energy which was give to an entirely different charge ? $\endgroup$ – Aditya Ahuja Dec 24 '19 at 14:36
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I know that the Q charge which gets stored on the capacitor comes from the connecting wires.

Just to be clear, charge is not stored in a capacitor. Charge is removed from one plate and an equal amount delivered to the other plate, for a net charge on each. It's the net charge on the plates that count because that's what creates the electric field between them.

However, since positive charge on one plate is reducing (Assuming conventional flow) and increasing on the other it is convenient to assume that the charge is going directly from one plate to the other as it makes calculating work done on the charges more easier.

The charge does not go directly from one plate to another. Although the same charge taken from one plate does not necessarily move through the battery and get delivered to the other plate, the battery has to supply energy (do work) in order for charge to be removed from one plate and charge delivered to the other plate, as discussed next.

Now, since it is not necessary that A went through the battery let us call the charge Q that went through the battery C. If battery didn't do work on A or B why do we say that the energy stored in the capacitor comes from the WORK done by the battery in transferring charges from one plate to the other.

Instead of thinking about the battery moving the same physical charge from one plate to the other, think about the battery separately taking charge from one plate at one of its terminals and delivering an equal amount of charge to the other plate from its other terminal, and supplying the necessary energy (work) in order to do this.

The battery needs to do work in order to put positive charge on the positive plate from its positive terminal, against the repulsive forces of the positive plate. And it needs to do work to remove an equal amount of positive charge from the negative plate against attractive forces of the negative plate. The energy needed to perform this work is the result of chemical reactions in the battery converting chemical energy to electrical potential energy, which winds up stored in the electric field of the capacitor.

Hope this helps.

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  • $\begingroup$ The electric field created by the battery pulls the charge from one plate and pushed the charge on the other plate. But the electric field in a wire is 0, right? So the entire work must be done inside the battery. So how does the battery pull and push charges on to the plate? $\endgroup$ – Aditya Ahuja Dec 21 '19 at 6:09
  • $\begingroup$ Also the charge only gains energy after going from the negative terminal to the positive terminal of the battery . It is easy to see that some work had to be on the charge done while getting PUSHED on to the capacitor plate. But how was work done on it to PULL it from the negative plate. Is work done by the field created by the battery equivalent to the work done by the battery ? (See my comment on The Photon's answer for a better explanation of my doubt ) $\endgroup$ – Aditya Ahuja Dec 25 '19 at 20:55
  • $\begingroup$ @AdityaAhuja What don't you understand about these answers (ALL of them)? They are answering your question very well. Work is done when charges are moved by a field with a direction component parallel to the field. That's the definition of work. Are you asking why charges experience a force when exposed to a field. There's no difference in "pushing" and "pulling" charges. $\endgroup$ – Bill N Dec 26 '19 at 4:31
  • $\begingroup$ @AdityaAhuja The e-field inside the wire is NOT zero. $\endgroup$ – Bill N Dec 26 '19 at 4:33
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    $\begingroup$ @Bill N Caution! Rabbit hole ahead! $\endgroup$ – Bob D Dec 26 '19 at 10:01

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