I see. You are talking about an infinitely large capacitor, but my example was a realistic case. The image below shows the electric field around a real capacitor.Remember that inside the ideal (conducting) wire, we have E→=0. So there is no electrostatic force present there. We also note that the electric field of the ideal capacitor (i.e. that produced by the charges which line the plates) is confined to the capacitor.
(Credit to Kogence)
If it was an ideal capacitor, you were right.
I don't know what you meant by this. I talked about the direction of the net force which has nothing to do with the battery. What I really mean is that you must do a bit work to counteract the electric field generated by the capacitor in reality. Can you elucidate you answer?The EMF is a line integral ∮(E→+1qf→chem)⋅dx→ around the whole circuit. The 1qf→chem exists only inside the battery, whilst the electric field (which is conservative, here) is not confined to the battery.
Inside the battery the electric field perfectly balances the chemical force (per unit charge). In an ideal wire, the electric field is zero. In a component like a resistor, we have a non-zero electric field which results in a voltage drop; the difference in energy here is that which is dissipated in the resistance.
Nevertheless, thank you for writing that equation. I just learned something new today.
I meant the energy released by the redox reaction occurring within the cells of the battery.I am not sure what the potential energy of a battery is . I don't know if this makes sense. But we can speak of the electric potential fields and chemical potential fields, for instance?