A capacitor consists of two parallel plates but one of them

A capacitor consists of two parallel plates, but one of them can move relative to the other as shown in the figure. Air fills the space between the plates, and the capacitance is 0.48 pF when the separation between plates is d = 0.500 cm. A battery supplying a potential difference V = 12.00 V is connected to the plates. What is the charge distribution, sigma, on the left plate? What are the capacitance, C\', and the charge distribution, sigma\', when d is changed to 0.250 cm? With d = 0.500 cm, the battery is disconnected from the plates. The plates are then moved so that d = 0.250 cm. What is the potential difference V\', between the plates?

Solution

sigma = Q / A. To find sigma, we need to know:
1. the charge on the capacitor (Q)
2. the area of the plates (A).

The charge on the capacitor: Q = CV = (.48x10^-12)(12) = 5.76x10^-12 C
The are of the plates: A = Cd / (epsilon-zero) = (.48x10^-12)(0.5x10^-2) / (8.85x10^-12) = 0.00027 m^2

sigma = Q / A = 5.76 * 10^-12 / 0.00027 = 2.13333x10^-8 C/m^2 <=== Answer

PART 2
When d is changed to 0.250 cm(decreases two times), the capacitance increases two times
The charge on the capacitor is doubled, and sigma is doubled
ANSWER: C\' = .96 pF, sigma\' = 4.2666x10^-8 C/m^2