Problem 20.45

20.45. a)
The circuit consists of two capacitors in parallel with a battery. The effective capacitance is given by
C_p=C₁+C₂
So, $C_{p} = 30 \mu F$
The energy stored in a capacitor is given by
$U = \frac{1}{2} C V^{2}$
So, in our problem,
$U_{p} = (.5)(30 \mu F)(100 V)^{2}~=~.150 J$

b)
Now, the two capacitors are connected in series with the battery. We want to find out what voltage is necessary to create the same energy, .150 J. First, we need to find the effective capacitance now that the capacitors are connected in series.
$C_{s} = \left(\frac{1}{C_{1}} + \frac{1}{C_{2}}\right)^{-1}$
Substituting in our values for C₁ and C₂, we find that
$C_{s} = 4.167 \mu F$
To find the potential difference necessary to maintain .150 J stored in the system, we solve our energy equation for potential.
$U = \frac{1}{2} C V^{2}$
$V = \sqrt{\frac{2U}{C}}$
$V = \sqrt{\frac{(2)(.15 J)}{4.167 \mu F}}$
V = 268 V