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Lower plate of a parallel plate capacito...

Lower plate of a parallel plate capacitor is fixed on a horizontal insulating surface. The upper plate is suspended above it using on elastic cord of force constant K. The upper plate has negligible mass and area of each plate is A. When there is no charge on the plates the equilibrium separation between them is d. When a potential difference = V is applied between the plates the equilibrium separation changes to x.
(a) Calculate V as a function of x.
(b) Find the value of x for which V is maximum. Calculate the maximum value of `V (= V_(max))`
(c) What will happen if `V gt V_(max)` ?
(d) Plot a rough graph showing variation of equilibrium separation (x) with V

Text Solution

Verified by Experts

The correct Answer is:
(a) `V=sqrt((2K(d-x)x^(2))/(in_(0)A))`
(b) `V_(max) = sqrt((8Kd^(3))/(27 in_(0)A)) "at" x=(2d)/(3)`
(c) System cannot remain in equilibrium
(d) `(##IJA_PHY_V02_C07_E01_048_A01##)`
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