Let us consider a battery with EMF E connected to an uncharged
capacitor and a resistance R. When the circuit is closed, an initial
current equal I=E/R is established in the circuit which drops
That is correct.
In a small time interval dt, a positive charge dq=I*dt will be
deposited on one plate of capacitor, which will attract same amount of
negative charge on the other plate and hence establish an electric
field between them.
The electric field is not established because the positive charge deposited on one platee "will attract the same amount of negative charge on the other plate". It is established because the positive terminal deposits positive charge one one plate while at the same time the negative terminal removes and equal amount of positive charge from the other plate. (What's really being deposited and removed are electrons, but that's another matter). In effect, the battery does work to separate the charge on the capacitor plates.
The electric field of battery doesn't do any work initially since the
capacitor is uncharged in the beginning.
Correct, because the voltage across the uncharged capacitor is zero. The potential difference $V$ between two points is defined as the work required per unit charge to move the charge between the two points. Since the initial voltage across the capacitor is zero, no work is initially required to move the charge. As the voltage increases, the work required increases. But the maximum work per unit charge the battery can do is its own emf, which is why charging stops when the capacitor voltage equals the emf of the battery.
I believe that later if battery adds more charge to the already
present charge, it will have to apply force against the electric field
of already deposited charges and thus do work in the process. Is my
Yes, because as the charge is separated the voltage increases, and thus the work required per unit charge to move more charge increases. In effect, it gets harder to pull positive charge off of a more negatively charged plate the more negative it becomes and to push positive charge onto a more positively charged plate the more positively it becomes, since the attraction and repulsive electrical forces increase.
However if I place a mechanical source of energy between plates of an
uncharged capacitor such as a conveyor belt (instead of a chemical one
which is a battery) which takes a positive charge dq from one plate
and deposits it on the other plate, it has to do work against the
electric field between positive and negative charge of magnitude dq to
create a charge separation.
It's a bit of an odd mechanical analogy, but OK. @Farcher has given you an excellent fluid mechanics analogy.
So how come a battery doesn't do any work to create an initial charge
separation of dq while a mechanical source of energy does? Are the two
processes of charging capacitors different or am I missing something
in the battery scenario?
I'm not sure why you think the conveyor does initially do work. Any mechanical work associated with moving the mass of the charge off and onto the plates would be negligible compared to the work against the force of the electric field.
In any event, like the battery, the conveyor also does not have to do any work to create the initial charge separation. That's because the attractive and repulsive forces that have to be overcome are initially negligible.
Hope this helps.