1. The problem statement, all variables and given/known data A 2.8 µF capacitor is fully charged to a value of 0.009 C. A 1.5 × 10^6 Ω resistor is connected to this capacitor so that it begins to discharge. One second after the capacitor has begun to discharge, how much energy is stored in the capacitor? Answer in units of J. 2. Relevant equations Q=CV Vo*(e^(-t/RC))=Vf 1/2C(Vf)^2 = E 3. The attempt at a solution .009C / 2.8*10^-6 F = 3214.28571429 V 3214.28571429V * e^-1 = 1182.4696323 V (1/2)* (2.8*10^-6 F) * (1182.4696323)^2 = 1.95752820396 J What I did was was divide .009 C by 2.8 * 10^-6 F to get the initial voltage then multiply by e^-1 to get the voltage in the capacitor. I then took that value, squared it, and multiplied it by 1/2 and 2.8 *10^-6 for the value of the capacitance. My final answer was 1.95752 J. I don't know what I did wrong.