1. The problem statement, all variables and given/known data Two parallel plate capacitors, 6 microfarads each, are connected in parallel to a 10V battery. One of the capacitors is then squeezed so that its plate separation is 50% of its initial value. Because of the squeezing, (A) how much additional charge is transferred to the capacitors by the battery and (B) what is the increase in the total charge stored on the capacitors? 2. Relevant equations C = ε A/d C = Q/V 3. The attempt at a solution I used C = ε A/d and found that the new capacity of the squeezed together capacitor is 12 microfarads. I then constructed an equivalent capacitor, with capacity of 18 microfarads and 10 V. I found that the charge (Q) here is (18 x 10^-6 Farads)(10V) = 1.8 x 10^-4 C. The original capacitor configuration, with equivalent capacity of 12 microfarads, could only hold 1.2 x 10^-4 C, so I found that the difference in charge was 1.8 x 10^-4 C - 1.2 x 10^-4 C, which equals 6 x 10^-5 C. I'm not sure whether this approach is right...and if it is, I don't know whether it's the answer to A or B. I can't tell the difference between what A is asking and what B is asking. If anyone could explain whether my approach is applicable to either A or B, and could also explain what the difference is between what both are asking, I would be super, super grateful. Thanks!