A parallel palte capacitor is filled by a dielectric whose relative permittively 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) = 78V`. It is then is connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.

Explanation

Assuming the requried final voltage be U. if C is the capacitance of the capacitor without the dielectric then the cahrge on the capacitor is given by `Q_(1)=CU` ltBrgt Since, the capacitor with the dielectric has a capacitance `epsiC`. Hence, the charge o the capacitor is given by
`Q_(2)=epsiCU=(alphaU)CU=alphaCU`
the initial charge on the capacitor is given by
`Q_(0)=CU_(0)`
From the conservation of charges, `Q_(0)=Q_(1)+Q_(2)`
Or `CU_(0)=CU+alphaCU^(2)`
`impliesalphaU^(2)+U-U_(0)=0`
`thereforeU=(-1+-sqrt(1+4alphaU_(0)))/(2alpha)`
On solving for `U_(0)=78V and a=2//V,` we get
`U=6V`.