Hard Capacitor Question AQA A Level Physics

IsaacJ2003

Initially a charged capacitor stores 1600 μJ of energy. When the pd across it decreases by 2.0 V, the energy stored by it becomes 400 μJ. What is the capacitance of this capacitor?

Im probably being stupid but this is a 1 mark question on an old aqa spec and the only way I can find of solving this. Is by creating the quadratic 0.25V^2 - 0.25v - 3 = 0 solving it and then placing it into 3200x10^-6 = CV^2 in order to find the capacitance. But this is meant to take at most 2 minutes in an exam, am I overlooking something super simple?

Reply 1

Ferret!

Capacitor energy = 0.5 * C * V²
=> (2E) / V² = C
Case 1:
(2 * 1600µ) / V² = C
Case 2:
(2 * 400µ) / (V - 2)² = C
A polynomial would come from the denominator, so you're doing the question correctly. A few multiple choice questions are just this tedious/time-consuming to do. :frown: