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

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SUMMARY

The discussion focuses on calculating the energy stored in a capacitor with a capacitance of 3000.0 μF charged to a voltage of 87.0 V, resulting in an energy storage of 6.53 J. The formula used for energy calculation is U = (1/2)C(dV^2), where C is the capacitance and dV is the voltage. The user incorrectly attempted to derive capacitance from energy and voltage, leading to confusion regarding the charge calculation using Q = C(dV). The correct approach maintains C as a constant value of 3000.0 μF for the initial question.

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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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