Hi there, new to this website so please bare with me if I break any rules and let me know for future. I am a maths and physics tutor but occasionally I come across a question I just cannot work the answer for. In this case I know the answer but I can't get to it. 1. The problem statement, all variables and given/known data Initially a charged capacitor stores 1600μJ of energy. When the pd across it decreases by 2.0V, the energy stored by it becomes 400μF. What is the capacitance of this capacitor? A. 100μF B. 200μF C. 400μF D. 600μF 2. Relevant equations E=(CV2)/2 rearrange the above to give C=2E/V2 3. The attempt at a solution So I've equated the two Cs to give me an equation that looks like E1/V2 = E2/(V-2)2 (the 2s have cancelled). I multiply each side by the opposite sides denominator and expand my squared bracket. Then by using the values of E given in the question I have a quadratic equation in terms of V that I can solve. This then gives me the values of V = 4 and V = 1.5. Using V = 4 with my E1 value I get the correct capacitance of 200μF. Do I know it must be V = 4 because I can't take 2volts away from V = 1.5? Is there a better way for me to do this? This part of the exam is multi choice and the average amount of time meant to be spent on each question is 90 seconds, and this way takes a wee while. Any thoughts about a better way to solve it would be appreciated as when I tried to do it a different way (I've thrown the sheet of paper away with that working on and can't remember how I did it) I kept getting the capacitance as 400μF. But I have been given the answer of B, 200μF, and using this value with values of E given and V = 4 and V = 2 on hyperphysics' capacitance stored applet I know that it must be 200μF. I'm tearing my hair out with this and really could do with some help! Thanks muchly, Steven Thomas.