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Physics
Classical Physics
Electromagnetism
Work done when Inserting a Dielectric between Capacitor Plates
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[QUOTE="kuruman, post: 6067130, member: 192687"] The disconnected case is simple because there is only kind of electrical force that does work on the system and that is due to the electric field generated by the isolated charge on the plates. In that case the work done by the electrical forces is ##dW=(-dU)## and the force would be ##F = -\left( \frac{\partial U}{\partial x} \right)_Q## where ##x## is the direction of insertion and the derivative is explicitly taken at constant ##Q##. The connected case is a bit more complicated because the work done by the battery needs to be taken into account. The work done on the system is ##dW=dW_{batt.}+(-dU)##. At constant voltage, when charge ##dQ## is added to the positive plate, ##dU=(1/2) V dQ##. Also, the work done by the battery to raise the potential energy of ##dQ## from zero to ##V## is ##dW_{batt}=VdQ##. Clearly then, ##dW_{batt}=2dU##. Thus the work done on the system at constant voltage is ##dW=2dU+(-dU)=+dU## and the associated force is ##F = +\left( \frac{\partial U}{\partial x} \right)_V## [/QUOTE]
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Physics
Classical Physics
Electromagnetism
Work done when Inserting a Dielectric between Capacitor Plates
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