Answer:
(a) 1.5 nF, 1.2 nF, 1 nF
(b) 0.4 nF
Explanation:
V = 150 V
V' = 40 V, V'' = 50 V, V''' = 60 V, q = 6 x 10^-8 C
(a) C' = q/V' = 6 x 10^-8 / 40 = 1.5 x 10^-9 F
C'' = q/V'' = 6 x 10^-8 / 50 = 1.2 x 10^-9 F
C''' = q/V''' = 6 x 10^-8 / 60 = 1 x 10^-9 F
(b) The effective capacitance is
[tex]\frac{1}{C}=\frac{1}{C'}+\frac{1}{C''}+\frac{1}{C'''}\\\\\frac{1}{C}=\frac{10^9}{1.5}+\frac{10^9}{1.2}+\frac{10^9}{1}\\\\C = 0.4\times 10^{-9} F[/tex]
