Energy of a Capacitor in the Presence of a Dielectric

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A dielectric-filled parallel-plate capacitor with a plate area of 20.0 m² and a separation of 5.00 m is connected to a 7.50 V battery, with a dielectric constant of 2.00. The energy of the capacitor needs to be calculated when it is half-filled with the dielectric. The relevant equation for energy in a capacitor is U = 1/2 CV², where C is the capacitance. The capacitance can be determined using the formula C = (κε₀A)/d, where ε₀ is 8.85×10−12 F/m. Understanding the units of area (m²) and separation (m) is crucial for accurate calculations.
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A dielectric-filled parallel-plate capacitor has plate area of 20.0 and plate separation of 5.00 . The capacitor is connected to a battery that creates a constant voltage of 7.50 . The dielectric constant is 2.00. Throughout the problem, use 8.85×10−12 .

The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy of the capacitor at the moment when the capacitor is half-filled with the dielectric.Express your answer numerically in joules.

I got confused ... and i am lost.. what equation should i use?
 
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plate area of 20.0 and plate separation of 5.00 .
What are the units of these quantities?
 
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