The circuit in the figure consists of two idential parallel metal plates connected by idential metal spring to a `50V` cell. With the switch open, the plates are uncharged, separated by a distance `d=10.0mm` and have capacitance `40muF`. When the switch is closed, the distance between the plates becomes half. (a) find charge on each plate.(b) find spring constant.

Explanation

(a) When switch is closed, separation becomes half i.e., capacitance is doubled.
`C=(Ain_(0))/(d)=40muF`
`C'=(Ain_(0))/(d//2)=(2Ain_(0))/(d)=80muF`
`Q=C'V=80xx10^(-6)xx50=4xx10^(-3)C`
(b) `(Ain_(0))/(d)=CimpliesAin_(0)=Cd`
Since separatiion becomes half i.e., `d` to `d//2`, extension of each spring `d//4`.
Consider equilibrium of one plate
`K(d)/(4)=F=(Q^(2))/(2Ain_(0))=(Q^(2))/(2cd)`
`K=(2Q^(2))/(Cd^(2))=(2xx(4xx10^(-3))^(2))/(40xx10^(-6)xx(10xx10^(-3))^(2))`
`=8000N//m`