Solve the all parts of given question:

Question: A parallel-plate capacitor has plates, each with an area of 4.0 cm², that are separated by a distance of 0.35 mm. The capacitor is first is connected to a 12 V battery, and fully charged. For parts (a) through (d) below, the battery is then disconnected.

Part I: What is the charge on the capacitor?

C

Part II: If the plates are then pulled to a separation of 0.95 mm, what is the charge on the capacitor now?

C

Part III: What is the potential difference across the plates now?

V

Part IV: How much work was required to pull the plates to their new separation?

J

Part V: Now go back to the first part of the problem with the capacitor connected to the battery and plate separation 0.35 mm. If the capacitor stays connected to the battery and the plates are pulled apart to a separation of 0.95 mm, what is the new potential difference across the capacitor?

V

Part VI: For the situation in part (e), what will be the new charge on the capacitor?

C

I need expert help to solve the above question.