- #1

- 41

- 1

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.