A parallel-plate air capacitor of area A= 21.0 cm2 and plate separation d= 3.20 mm is charged by a battery to a voltage 50.0 V. If a dielectric material with κ = 4.80 is inserted so that it fills the volume between the plates (with the capacitor still connected to the battery), how much additional charge will flow from the battery onto the positive plate? C= ε0A/d Q=C(dV) C= κC_0_ Q= κQ_0_ Using the first equation, I found the capacitance for the vacuum-insulated capacitor. Then I found the change in capacitance using C=κC_0_. I got C_0_(vacuum-insulated capacitance) to be 7.29e-9 F and got C(dielectric-insulated capacitance) to 3.5e-8. I know that while the capacitor is hooked up to the battery, the potential difference(dV) does not change. So I used Q= C(dV) to find the charge on the plates with dielectric in between which was 1.75e-6 C. I also found the Q_0_(vacuum-insulated charge) to be 3.64e-7 C. I took the difference of the two and got 1.38e-6 C. I thought this was the answer but it was wrong. The problem asks, "how much additional charge will flow from the battery onto the positive plate?" though. I feel I'm missing something but I don't know what.