A plate A of a parallel-plate capacitor is fixed, while a plate B is attached to the wall by a spring
Answers
Since the battery remains connected, V remains constant. C decreases as d increases (C ∝ 1/d) and UC = ½ CV2
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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 plate A (Fig. 97). After 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 plate B cannot be shifted noticeably?
Concept:
- Parallel plate capacitors
- The voltage across the capacitor is kept constant and equal to the battery's emf while key K is closed.
Given:
- When the new equilibrium point is reached, let the displacement of plate B be -x1.
- Let area of capacitor plates be S
- Rigidity of spring = k
Find:
- the decrease in the equilibrium separation between the plates
Solution:
In this instance, the capacitor is charged at q1 = ξC1 = εSξ/ (d - x1).
The capacitor has a field strength of E1 = ξ/(d - x1), but it is created by two plates.
As a result, the field strength created by one plate is E1/2, and we may write the force acting on plate B as
E1q1/2 = εSξ²/2 (d - x1)² = kx1
Let's now explore the scenario where the key K is temporarily closed.
The capacitor builds up a charge q2 = εSξ/d (the plates can't move) that doesn't change.
Let x2 be the plate B's displacement at its new equilibrium position.
The capacitor's field strength is then given by E2 = q2/[C2 (d - x2)] and C2 = εS/ (d - x2). The equilibrium circumstance for plate B in this instance can be expressed mathematically as
E2q2/2 = εSξ²/2d² = kx2
Dividing the first equation by the first, we get x1 = 0.1d and x2 = 0.08d
The decrease in equilibrium separation is 0.02d.
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