Solve the all parts of given question:
Question: A parallel-plate capacitor with plate area 4.0cm2 and air-gap separation 0.50 mm is connected to a 9.0-V battery, and fully charged. The battery is then disconnected.
Part A: What is the charge on the capacitor?
Part B: The plates are now pulled to a separation of 0.75 mm. What is the charge on the capacitor now?
Part C: What is the potential difference between the plates now?
Part D: How much work was required to pull the plates to their new separation?
Please include each step and the equations you used. Thanks!