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?