In one type of computer keyboard, each key holds a small metal plate that serves as one plate of a parallel$-$plate, air$-$filled

capacitor. When the key is depressed, the plate separation decreases and the capacitance increases. Electronic circuitry

detects the change in capacitance and thus detects that the key has been pressed. In one particular keyboard, the area of

each metal plate is $52\mathrm{.}0m{m}^2,$ and the separation between the plates is $0\mathrm{.}660mm$ before the key is depressed.

Part A

Calculate the capacitance before the key is depressed.

Express your answer in farads.

Part B

If the circuitry can detect a change in capacitance of $0\mathrm{.}300pF,$ how far must the key be depressed before the circuitry detects its depression?

Express your answer in millimeters.