Dielectric medium: relative permittivity ε = α U
so permittivity : ε ε₀ = α U ε₀
where U = applied voltage and α = 2 / volt
Capacitance C₁ = Q / V = ε ε₀ A / d = 2 U ε₀ A / d
Capacitance C₂ = ε₀ A / d
C₁ = C₂ * 2 U
Given, V₂ = U₀ = 78 Volts.
Initially, the Capacitor C1 with the dielectric medium is not charged. Hence,
Q₁ = V₁ * C₁ = 0
The charge on capacitor C₂ : Q₂ = C₂ V₂ = C₂ * 78 Coulombs
When the two capacitances are connected in parallel, the charges are distributed among the both. The equivalent capacitance is C = C₁ + C₂. The voltage across both capacitors will be same.
C = C₂ (1 + 2 U)
Q = charge on both capacitors = Q₂ = 78 C₂
Final voltage across both : U = Q / C = 78 / (1 + 2 U)
=> U + 2 U² - 78 = 0
=> U = [ -1 + - √(1 + 624) ] / 4
=> U = 6 Volts.
So the capacitance of C₁ = 12 C₂ Farads and potential difference is 6 Volts.