[tex]\Delta [/tex]Uand W, increasing the separation of capacitor, with V=const.?

[Monnn]

I expected it to be quite simple as it was without the battery, in which case $$\Delta$$U = W, but now I am confused.

with the separation increased from x to x+$$\Delta$$x,

It is obvious that $$\Delta$$U changes with 0.5[1/x - 1/(x+$$\Delta$$x)] since C changes with 1/r. (Without the common coefficient, Q^2 over A epsilon0 or something)

And the battery which maintains the voltage difference between the plates gains energy from the system with V$$\Delta$$Q = [1/x - 1/(x+$$\Delta$$x)] (Note that it is without 0.5 which sticks with the energy difference)

And the work exerted on the system by moving the plates is (pressure)*(area)*(distance) an integrand over 1/r^2, which gives 0.5[1/x - 1/(x+$$\Delta$$x)] as well (the factor 0.5 originates from that of the electrostatic pressure).

So, 0.5 is exerted on the system and it seems that it is stored in the system, as $$\Delta$$U. Then how should I explain the work with the battery?

 

[jbsteele]

Is it the same 0.5[1/x - 1/(x+\Deltax)] or is it something else?The work done by the battery is equal to the energy stored in the system, which is \DeltaU = 0.5[1/x - 1/(x+\Deltax)]. The battery provides energy to the system to maintain the voltage difference between the plates, and this energy is stored in the system as potential energy.