Two parallel plate capacitors with area `A` are connected through a conducting spring of natural length `1` in series as shown. Plates `P` and save fixed positions at separation `d`. Now the plates are connected by a battery of `emf xi` as shown. If the extension in the spring in equilibrium is equal to the separation between the plates, find the spring constant `k`.
.

Options

A. `k=(27)/(8)(A epsilon_(0) xi^(2))/((d-l)^(3))`
B. `k=(27)/(8)(A epsilon_(0) xi)/((d-l)^(3))`
C. `k=(28)/(8)(A epsilon_(0) xi)/((d-l)^(3))`
D. `k=(27)/(8)(A epsilon_(0) xi)/((d-l)^(2))`

Answer

A

Explanation

Sol : Let charge on capacitors be `q` and separation between plates `P` & `Q` and `R` & `S` be `x` at any time. Distance between pates `P` & `Q` and `R` & `S` is same because force acting on them is same.

Capacitance of capacitor `PQ,C_(1)=(epsilon_(0)A)/(x)`
Capacitance of capacitor `RS,= (epsilon_(0)A)/(x)`
From `KVL`, we have `(1)/(C_(1))+(q)/(C_(2))=xi or q=(epsilon_(0)A xi)/(2x)`
Force on plates `Q` towards `P`
`F_(1)=(q^(2))/(2A epsilon_(0))=(epsilon_(0)A^(2) xi^(2))/(8 Ax^(2)epsilon_(0))=(A epsilon_(0) xi^(2))/(8x^(2))`
Spring force on plate `Q` due to extension in spring At equilibrium, separation between plates = extension is spring
Thus `x=y(-d-2x-l)rArr x=(d-l)/(3)` ....(1) ltbrlt At equilibrium, `F_(1) = F_(2)` ....(2)
From equation (1) and (2),we have
`(A epsilon_(0) xi^(2))/(8x^(2))=ky=kx rArr x=((A epsilon_(0) xi^(2))/(8k))`....(3)
From equation (1) and (3),
`((d-1)/(3))^(3)=(A epsilon_(0) xi^(2))/(8k)rArr k=(27)/(8)(A epsilon_(0) xi^(2))/((d-l)^(3))`.