Here I have a parallel plate capacitor connected to a battery with emf V. Now I pull the plates of the capacitor to increase its plate distance. As I do some work, this work appears as the increase in potential energy between the plates.
My doubt is, as there is increase in potential difference between the plates which is different than the potential difference intially ie, V. How does it persist? Battery's job is to maintain same V but what happens in the circuit (in terms of potential) due to increasing d.
1this work appears as the increase in potential energy between the plates.
Work done appears as the decrease in electrostatic potential energy as the capacitance decreases.
as there is increase in potential difference between the plates which is different than the potential difference intially
Prove your point.
1When I increase the plate seperation, shouldn't the potential increase by the relation V=Ed . (charge isn't changing here implying there is no change in electric field between the plates).
The battery suppose maintains a potential V b/w the plates of the capacitors, But as I increase the seperation, the potential should differ from the pt above. How does this balance.
As you said capacitance decreases, will some charge flow out of the plates to the battery?
Thanks
1When you decrease the capacitance, some charge (Q) flows into the battery from the capacitor.
V=E.d
Note that E depends on the charge density. Since Q decreases, E is ought to decrease.
You can prove doing some math that this decrease in E compensates the increase in d such that V remains constant.