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?(adsbygoogle = window.adsbygoogle || []).push({});

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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# Homework Help: Help appreciated

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