Two identical capacitors are connected parallel. Initially they are charged to a potential V₀ and each acquired a charge Q₀. The battery is then disconnected, and the gap between the plates of one capacitor is filled with a dielectric κ.(a) What is the new potential difference V across the capacitors. possible asnwers: V=(V₀)^2/[kQ₀+V₀), V=V₀/2k, V=V₀/2, V=kQ₀/V₀, V=2V₀/[k+1](b) If the dielectric constant is 7.8, calculate the ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge.

ΔV ’= 2Δv / (1 + k) ,            Q’= Q₀ 1,772

Explanation:

When the capacitors are connected in series the capacitance is added

      C_eq = C₁ + C₂2

     C_eq = 2 C₀

     ΔV = Q₀ / C_eq

     ΔV = Q₀ / 2 C₀

In this case, a dielectric is introduced to one of the capacitors, so its capacity changes.

         C₂’= k C₂

Since the two capacitors have the same initial value let's call Co

          C_eq’= C₀ (1 + k)

The set charge is

           C_eq ’= Q₀ / ΔV’

            ΔV ’= Q₀ / Ceq’

            ΔV ’= Q₀ / C₀ (1 + k)

The relationship between these voltages is

            ΔV'/ ΔV = 2 / (1 + k)

            ΔV ’= 2Δv / (1 + k)

The initial charge of the capacitor is

              Q₀ = C₀ ΔV

The charge after introducing the dielectric is

             Q’= k C₀ ΔV’

             Q ’= k C₀ 2 ΔV / (1 + k)

             Q’= Q₀ 2k / (1 + k)

             Q’= Q₀  2 7.8 /(1+7.8)

              Q’= Q₀ 1,772

the gas collects in coils that then becomes a liquid that then flows through the coils inside the refrigerator

i think the answer would be b