The arrangement of parallel conducting plates shown in are of the same surgface area `A=10 cm^(2)`. A battery of `E=10V` is connected aross the kends `A` and `B`. Plate 2 is slowly moved upward by some exteral force.
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find the position of plate at which the energy stored in the system is minimum. Also, find this mimimum energy Take `epsilon_(0)=8.8xx10^(-12)Fm^(-1)`.

Answer

`2 cm` from plate-3, `4.4xx10^(-13) J`

Explanation

Let `x` the instantaneous distance distance between the plate `2` and `3`, then that between `1` and `2` will be `d-x`, where `d=3 cm`. Also, distance between `3` and `4` is `d_(0)=1 cm`.
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The system of plates can be represented as combibination of three capacitors `C_(1), C_(2)` and `C_(3)`, where
`C_(1)=(epsilonA)/(d-x),C_(2)=(epsilon_(0)A)/x` and `C_(3)(epsilon_(0)A)/d_(0)`

The equivalent capacitance of the system is
`C=(C_(1)C_(3))/(C_(1)+C_(3)+C_(2))=(epsilon_(0)A)/(d_(0)+d-x)=(epsilon_(0)A)/x`
The energy of the system is minimum when
`(dC)/(dx)=0`.
`:. (dC)/(dx)=(epsilon_(0)A)/(d_(0)+d-x)^(2)-(epsilon_(0)A)/x^(2)=0`
or `x^(2)-(d+d_(0)-x)^(2)=0` or `x=(d+d_(0))/2=(3+1)/2=2 cm`
The minimum capacitance is `C_(m)=epsilon_(0)A`
and the mimimum energy is
`U_(m)1/2C_(m)E^(2)=1/2epsilon_(0)AE^(2)`
`=1/2(8.8xx10^(-12))(10xx10^(-4))(10)^(2)=4.4xx10^(13)J`.