A parallel-plate capacitor has a charge Q and plates of area A. What force acts on one plate to attract it toward the other plate? Because the electric field between the plates is E = Q /A) 0, you might think that the force is F = QE = Q2/A) 0. This is wrong, because the field E includes contributions from both plates, and the field created by the positive plate cannot exert any force on the positive plate. Show that the force exerted on each plate is actually F = Q2/2)0A. (Suggestion: Let C =) 0A/x for an arbitrary plate separation x; then require that the work done in separating the two charged plates be W =) F dx.) The force exerted by one charged plate on another is sometimes used in a machine shop to hold a work piece stationary.