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This is the text from Griffiths Electrodynamics for calculating force on dielectric on pulling it out from capacitor when the capacitor is connected to Battery so that voltage of capacitor is fixed:

.... The work done on dielectric is :

$$ dW= F_{me} dx+ VdQ$$

Where W is work done on dielectric ,F is the force is Applied,VdQ is work done by Battery....

The question is how the last term came up(VdQ)? Which Q (bound or free) is he referring to? How exactly the battery does work here?

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Not seeing the equation in context my guess would be the following:

$VdQ$ is the differential amount of work a battery of potential $V$ would do in putting a differential amount of charge on the capacitor plates in the first place thereby storing energy in the electric field of the capacitor. So the work done to remove the dielectric would be the mechanical work to physically remove the material plus the equivalent of the electrical work to take the charge off the plates equal to the electrical work that was required to put the charge on the plates.

Hope this helps.

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