A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U ) as `epsilon = alpha` U where`alpha=2V^(-1)`. A similar capacitor with no dielectric is charged to `U_(0)` = 78 V. lt is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.

Explanation

Let the capacitance of capacitor without dielectric is C and hence, charge on capacitor
`Q_(1)=CU`
where U is the final potential of a capacitor
As the capacitor with dielectric having relative permittivity `epsilon` and its capacitance `in` C. Hence, charge on the capacitor is,
`Q_(2)= in CU = aUxx CU= aCU^(2)`
`[ because in = aU]`
The initial charge on the capacitor is,
`Q_(0)=CU_(0)`
From the conservation of charge,
`Q_(0)= Q_(1)+Q_(2)`
`CU_(0)=CU+aCU^(2)`
`:.aU^(2)+U-U_(0)=0`
Now a = `2V^(-1)` and `U_(0)` = 78 V
`:. 2U^(2) + U-78=0`
is a binomial equation of U.
`:. 2U^(2) +13U -12 U -78 =0`
`:. U(2U+13) - 6 (2U+13) =0`
`:.(2U+13)` (U-6) =0
`:. U= -(13)/(2) "or" U =6`
`:. U = -(13)/(2)` is impossible
`:. U = 6V` final voltage