How should four 2.0 uF capacitors be connected to have a total capacitance of a.

Question

How should four 2.0 uF capacitors be connected to have a total capacitance of a. 8.0 uF? b. 2.0 uF? c 1.5 uF? d. 0.5 uF?
I was able to figure out a with the help of the instructors notes.
C= C1 + C2 + C3 + C4 = 8.09 uF

b. is two connected in parallel, followed by two more connected in parallel. It sounds like they are saying add two together and then put them in series? So 2 + 2 = 4 and then I put that in series.

C = 1/4 + 1/4 = 2/4 = 0.5, except the answer is 2 uF. Not sure where to go with it from here. Thanks!
2nd posting, since the last time, two people reposted the answer I provided. I'm asking why and how, not what. I know it is parallel, that's the answer. I can't make it work out. Thanks again!

Explanation / Answer

Energy is proportional to capacitance when connected in parallel to the same voltage source. So energy stored by 4uF is twicw that of 2uF and that of 6uF is three times that of 2 uF. When connected in series all of them store the same charge q = V C(eq); C(eq) = 1/[1/2+1/4+1/6] = 12 /11 uF so q = 12V/11 The energy stored by 2uF would be (1/2q)(q/2). This shows that 2uF will store three times more energy thn 6 uF and 4 uuF will store 1/2 of what 2u F stores. So in first case energy stored would be 1:2:3 and in seriescase 3:2:1. It also follows from the expression for energy, 1/2) Cv^2, (1/2) (Q^2)/C and (1/2)QV

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