Calculating Work Required to Remove Dielectric Slab from Charged Capacitor

[srhly]

An isolated capacitor has a dielectric slab k between its two plates. The capacitor is charged by a battery. After the capacitor is charged, the battery is removed. The dielectric slab is then removed. Finally, the capacitor reaches equilibrium. Initially the parallel-plate capacitor has a capacitance of 2.19nF is charged to an initial potential difference of 94 V. The dielectric material has a dielectric constant k = 7.54. What is the magnitude of the work required when the dielectric slab is then removed?

I used:
W= Ui - Uf
U= 0.5CV^2
Uf= (0.5)C(k*Vi)^2
Ui = (0.5)C(Vi)^2

Vi=initial potential difference.

I have tried to double check the math, so I guess I messed up with the formulas. Where did I go wrong?

 

[mukundpa]

After charging, the battery is removed hence by displacing the slab the capacitance will change produces change in potential difference between the plates.

In such processes where the battery is disconnected the charge will remain unaltered, not potential difference.