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A plate A of a parallel-plate capacitor is fixed, while a plate B is attached to the wall by a spring and can move, remaining parallel to the plate A (Fig. 97). After the key K is closed, the plate B starts moving and comes to rest in a new equilibrium position. The initial equilibrium separation d between the plates decreases in this case by 10%.

What will be the decrease in the equilibrium separation between the plates if the key K is closed for such a short time that the plate B cannot be shifted noticeably?

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When the key K is closed, the voltage across the capacitor is maintained constant and equal to the emf ξ of the battery. Let the displacement of the plate B upon the attainment of the new equilibrium position be -x1. In this case, the charge on the capacitor is q1 = C1ξ = ε0Sξ/(d - x1), where S is the area of the capacitor plates. The field strength in the capacitor is E1ξ/(d - x1), but it is produced by two plates. Therefore, the field strength produced by one plate is E1/2, and for the force acting on the plate B we can write

where k is the rigidity of the spring.

Let us now consider the case when the key K is closed for a short time. The capacitor acquires a charge q2 = ε0Sξ/d (the plates have no time to shift), which remains unchanged. Let the displacement of the plate B in the new equilibrium position be x2. Then the field strength the capacitor become E2 = q2/[C2 (d - x2)] and C2ε0S/(d - x2). In this case, the equilibrium condition for the plate B can be written in the form

Dividing Eqs. (1) and (2) term wise, we obtain x2 = x1 [(d - x1)/d]2. Considering that x1 = 0.1d by hypothesis, we get x2 = 0.08d.

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