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

On connecting the two given capacitors,let the final voltage be U.
If capacity of capacitor of capacitor without the dielectric is C, then the charge on this capacitor is `Q_(1) = CU`
The other capacitor with dielectric has capacity `in C` . Therefore, charge on it is `Q_(2) = in CU`.
As `in = alpha U`, therefore, `Q_(2) = alpha C U^(2)`
The initial charege on the capacitor (without the dielectric is C, then the charge on this capacitor is `Q_(1) = CU`
The other capacitor with dielectric has capacity `in C`. Therefore, charge on it is `Q_(2) = in CU`.
AS `in = alpha U`, therefore, `Q_(2) = alpha C U^(2)`
The initial charge on the capacitor (without dielectric) that was charged si `Q_(0) = CU_(0)`
From the conservation of charge, `Q_(0) = Q_(1) + Q_(2)`
`CU_(0) = CU + alpha CU^(2) or alpha U^(2) + U - U_(0) = 0 :. U = (-1 +- sqrt(1+4 alpha U_(0)))/(2 alpha)`
Using `alpha = 2 V^(-1) and U_(0) = 78 vol t`, we got `U = (-1 +- sqrt(1+4xx2xx78))/(2xx2) = (-1 +- sqrt(625))/(4)`
As U is positive, therefore , `U = (sqrt(625) -1)/(4) = (24)/(4) = 6vol t`