A paallel plate capacitor with air as a dielectric is arranged horizontally. The lower plate is fixed and the other connected with a perpendicular spring. The area of each plate is `A` and mass of uppar plate is `m`. In the steady position, the distance between the plates is `d_(0)`. When the capacitor is connected with a battery of emf `V`, a new equilibrium appears, with distance between the plate as `d`.
(a) find the spring constant `k`.
(b) what is the maximum voltage for a given `k` in which an equilibrium is possible?

Explanation

(a) Consider equilibrium of upper plate
when capacitor is connected to battery, battery supplies charge to plates and plates attract each other.
`F=(Q^(2))/(2Ain_(0))=(C^(2)V^(2))/(2Ain_(0))=((Ain_(0))/(d))^(2)(V^(2))/(2Ain_(0))`
`=(1)/(2)(Ain_(0)V^(2))/(d^(2)`
Further extension of spring `=(d_(0)-d)`
Again equilibrium of upper plate
`k(x_(0)+d_(0)-d)=mg+F=mg+(1)/(2)(Ain_(0)V^(2))/(d^(2))`
`k=(Ain_(0)V^(2))/(2d^(2)(d_(0)-d))`
(b) `V^(2)=(2kd^(2)(d_(0)-d))/(Ain_(0))=(2k(d^(2)d_(0)-d^(3)))/(Ain_(0))`
For `V` to be maximum, `(dV)/(dd) or (dV^(2))/(dd)=0`
`(dV^(2))/(dd)=2dd_(0)-3d^(2)=0impliesd=(2d_(0))/(3)`
`V_(max)^(2)=(2k[((2d_(0))/(3))^(2)d_(0)-((2d_(0))/(3))^(3)])/(Ain_(0))`
`=(2kd_(0)^(3))/(Ain_(0))[(4)/(9)-(8)/(27)]=(2kd_(0)^(3))/(Ain_(0))((4)/(27))`
`=(8kd_(0)^(3))/(27Ain_(0))`
`V_(max)=sqrt((8kd_(0)^(3))/(27Ain_(0)))`