Calculating Work Required to Remove Dielectric Slab from Charged Capacitor

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To calculate the work required to remove a dielectric slab from a charged capacitor, the initial and final energy states must be determined using the formulas W = Ui - Uf, where Ui is the initial energy and Uf is the final energy. The initial energy is calculated with Ui = 0.5 * C * (Vi)^2, and the final energy after removing the dielectric is Uf = 0.5 * C * (k * Vi)^2. The discussion highlights that after the battery is removed, the charge remains constant while the potential difference changes due to the removal of the dielectric. The confusion arises from incorrectly applying the formulas, as the change in capacitance affects the potential difference rather than the charge. Understanding these principles is crucial for accurately calculating the work involved in this process.
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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?
 
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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.
 

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