Parallel-Plate Vacuum Capacitor

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When a parallel-plate vacuum capacitor is disconnected from its potential source and the plate separation decreases from x1 to x2, the energy stored in the capacitor changes. The charge (Q) remains constant since the capacitor is isolated. The energy can be expressed using the formula U = Q^2 / (2C), where C is the capacitance that varies with the plate separation. The capacitance for each configuration can be calculated, allowing for the determination of the new energy stored after the separation adjustment. Thus, the energy stored in the capacitor after the change in plate separation can be accurately calculated using these principles.
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If a parallel-plate vacuum capacitor, with its plates separated by a distance of x1 has an amount of energy equal to U stored in it. The separation is then decreased to x2 .

What is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed? I know that I am suppose to use the formula C= Q/V but how do I put U in terms of x1 and x2?
 
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As the battery is disconnected, the charge on the capacitor plates becomes constant. Now by changing capacitance the potential difference will change and hence it is batter to use the formula Q^2 /2C of the energy.

Capacitance in the two situations are to be calculated and Q is calculated by first situation only.
 

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