On connecting the two given capacitors, let the final voltage be V.
If capacity of capacitor without the dielectric is C, then the charge on this capacitor is q1=CV
The other capacitor with dielectric has capacity εC.
Therefore, charge on it is q2=εCV
As ε=αV, therefore q2=ε=αV
The initial charge on the capacitor (Without dielectric)
that was charged is q0=CV0
From the conservation of charge, q0=q1+q2
CV0=CV+αCV2orV2+V−V0=0
V=−1±√1+4αV02α
using α=2V−1 or V0=78V, we get
V=−1±√1+(4×2×78)2×2=−1±√6254
As V is positive, therefore, V=√625−14=244=6V,