The potential difference in a capacitor with dielectric

[Ripperbat]

Homework Statement

Two identical capacitors are connected in parallel, charged by a battery ov Voltage V, and then isolated from the battery. One of the capacitors is then filled with a material of relative permittivety E.

Homework Equations

What is the final potential difference across it's plates?

The Attempt at a Solution

Potential difference?

Q = Charge

2C in parallel

Q₁ = C₁V
Q₂ = C₂V

C₁=C₂=C

Q=Q₁+Q₂ = CV+CV=2CV (on each capacitor)

Q=2C₀V -> C₀ =( Q / (2V) )

The battery is removed, one capacitor is filled with dielectric with relative permittivety E.

C = EC₀

Q = C_fV_f = C_iV_i

Vf = (C_i/C_f)*Vi = (C_iV_i)/C_f = (2C₀V)/EC₀ = 2V/E

The correct solution should look like: (2V)/(E+1)

 

[ehild]

The sum of the charge on both capacitors stays the same as before: Q=2VCo, but the new resultant capacitance is (E+1)Co, so the new voltage is V'= Q/Cresultant= 2VCo/((E+1)Co)

ehild

 

[Ripperbat]

The sum of the charge on both capacitors stays the same as before: Q=2VCo, but the new resultant capacitance is (E+1)Co, so the new voltage is V'= Q/Cresultant= 2VCo/((E+1)Co)

ehild

So what they mean by isolating the circuit from the battery is that it still remains a closed circuit, hence we have to take both capacitors into account when solving this?