Question
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An electrostatic actuator can be made by placing a movable dielectric inside a parallel plate capacitor. The presence of charge on the plates will attract the movable dielectric. Assume that a parallel plate capacitor with area A and separation d is initially charged to a voltage V by an external battery. The battery is then removed so that a charge Q is left on the positive plate and -Q is left on the negative plate. Assume also, that an electric field exists only between the plates and there are no fringe fields. A dielectric slab with area A and thickness d is inserted slowly into the parallel plate capacitor. Find the potential energy stored in the capacitor before the dielectric is inserted. Find the voltage on the capacitor when the dielectric is inserted half-way. Find the voltage on the capacitor when the dielectric is fully inserted. Find the potential energy stored in the capacitor after the dielectric is fully inserted. Find the force required to displace the dielectric from the fully inserted position.
Explanation / Answer
C=Aepsiolon/d
1)Q^2/2C
2)Q/(k+1)/2C
3)Q/KC
4)Q^2/KC
5)U=Q^2/(kx+(1-x)))C
F=-dU/dX
=0
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