Dielectric inserted into a parallel plate capacitor

[Vriska]

Homework Statement

The capacitor (of thickness d) is disconnected from a potential source of V and a dielectric of thickness t is inserted and it has relative permitivity Er. Find the new potential between the plates

Homework Equations

[/B]
This is the answer : Vf = V/d(d - t + t/Er)

The Attempt at a Solution

I don't get why this is it looks like they assumed that the electric field would remain constant and just added the potential drops. We know electric field isn't constant so what give?

 

[Baluncore]

The charge is isolated and so is the only thing that remains constant.
The voltage changes as a result of the changing capacitance. C = Q / V.