Answer
The maximum energy that can be stored in the three-capacitor arrangement is $~~0.095~J$
Work Step by Step
In part (a), we found the potential difference across each capacitor in the maximum arrangement:
$V_1 = 100~V$
$V_2 = 50.0~V$
$V_3 = 40.0~V$
We can find the energy stored in each capacitor:
$U_1 = \frac{1}{2}(10.0\times 10^{-6}~F)(100~V)^2 = 0.050~J$
$U_2 = \frac{1}{2}(20.0\times 10^{-6}~F)(50.0~V)^2 = 0.025~J$
$U_3 = \frac{1}{2}(25.0\times 10^{-6}~F)(40.0~V)^2 = 0.020~J$
We can find the total stored energy:
$U = U_1+U_2+U_3$
$U= 0.050~J+0.025~J+0.020~J$
$U = 0.095~J$
The maximum energy that can be stored in the three-capacitor arrangement is $~~0.095~J$.
