For the capacitor network shown in the figure , the potential difference acrossabis 36V. Determine:(a) the total charge stored in this network,(b) the charge on each capacitor,(c) the total energy stored in the network,(d) the energy stored in each capacitor,(e) the potential difference across each capacitor.We have two capacitors in series here.Looking from left to right in the figure, we’ll have chargesof +Qand-Qon the first capacitor’s plates, and charges of +Qand-Qon the second capacitor’splates. These two in series make some single equivalent capacitor with the sameQand-Qacross itsplates. Since they are in series:1C=1C1+1C2. A little algebra lets us write this asC=C1C2C1+C2so hereC= (150nF)(120nF)/(150nF+ 120nF) = 66.67nF.(a) Now that we know the completeCfor this network,C=Q/VsoQ=CV= (66.67×10-9)(36.0) =2.40×10-6Coulor 2.40μC.(b) From the first paragraph, for these capacitors in series, we have the sameQas we just calculatedon each capacitor.(c) The total energy stored isU=12CV2= (0.5)(66.67×10-9)(36.0)2= 43.2×10-6Jor 43.2μJ.(d) To find the energy stored in each capacitor individually, we have various forms forU.At thismoment, we don’t (yet) know the voltage drop across each capacitor separately, so one option is touseU=Q2/(2C) since we DO know the charge and capacitance of each capacitor.For the first capacitor:U= (2.40×10-6)2/(2×150×10-9) = 1.92×10-5Jor 19.2μJ.For the second capacitor:U= (2.40×10-6)2/(2×120×10-9) = 2.40×10-5Jor 24.0μJ.(Adding these together, we get 43.2μJ, which is the same energy we calculated was stored in theentire network, in part (c).)(e) To find the potential drop (i.e.the voltage) across each capacitor individually,C=Q/VsoV=Q/Cand we do know the charge and capacitance of each capacitor. The voltage drop across thefirst capacitor will be:V= (2.40×10-6C)/(150.0×10-9F) = 16.0V. The voltage drop across thesecond capacitor will be:V= (2.40×10-6C)/(120.0×10-9F) = 20.0V. (And as a check, the voltagedrop across the two capacitors combined would be 16 + 20 = 36Vwhich is just what we were told itwas.)

Capacitor in a computer keyboardIn one type of computer keyboard, each key holds a small metal plate that serves as one plate of aparallel-plate, air-filled capacitor. When the key is depressed, the plate separation decreases and thecapacitance increases. Electronic circuitry detects this change in capacitance and thus detects thatthe key has been pressed. In one particular keyboard, the area of each metal plate is 42.0mm2, andthe separation between the plates is 0.700mmbefore the key is depressed.(a) Calculate the capacitance before the key is depressed.