Calculate Capacitor Energy: 3000.0 μF, 87.0 V & 6.53 J | Formulas & Solutions

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The discussion focuses on calculating the energy stored in a capacitor with a capacitance of 3000.0 μF charged to 87.0 V. The energy formula U = (1/2)C(dV^2) is used to find the energy, which results in 6.53 J. However, confusion arises when attempting to calculate the charge using the same capacitance value for a different energy scenario. The correct approach is to recognize that capacitance (C) is constant, while the voltage (dV) changes for different energy calculations. The error stems from misapplying the capacitance value in the charge calculation, as it should remain constant at 3000.0 μF.
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A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electrical capacity of 3000.0 μF and is charged to a potential difference of 87.0 V. Calculate the amount of energy stored in the capacitor.

Calculate the charge on this capacitor when the electrical energy stored in the capacitor is 6.53 J.

U= (1/2)C(dV^2)
Q= C(dV)

I found the first answer but for the 2nd part with the 6.53 J, I'm not getting the right answer.
I figured I'd use U= (1/2)C(dV^2) and solve for C. Then plug C into Q= C(dV). dV is the same for both questions.

U= (1/2)C(dV^2)
2U/(dV^2)= C
C= 1.72e-3 F
Then,

Q= C(dV)= .14964 C
That's not right answer though. What am I doing wrong?
 
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You already know C it is given by the question-> 3000.0 μF
C is a constant not V; V will be different for second question.
 
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