**We charge a capacitor of capacitance $C_{1} = 8.0 \, \mu F$ by connecting it to a source of potential difference $V_{0} = 120 \, V$. Once C_{1} is charged, the potential difference is disconnected.**

**What is the charge $Q_{0}$ on $C_{1}$?****What is the energy stored in $C_{1}$****A second uncharged capacitor of capacitance $C_{2} = 4.0 \,\mu F$ is connected in parallel to $C_{1}$. What is the potential difference across each capacitor, and what is the charge on each capacitor? What is the total energy of the system?**

**1.**

$$\begin{aligned} Q_{0} &= \frac{V}{C} \\ &= \frac{120}{8.0 \mu} \\ &= 960 \, \mu F\end{aligned}$$

**2.**

$$\begin{aligned} U &= \frac{1}{2} Q_{0} V \\ &= \frac{1}{2} (960 \mu) (120) \\ &= 0.058 \, J \end{aligned}$$

**3.**

The equivalent capacitance is:

$$\begin{aligned} C_{eq} &= C_{1} + C_{2} \\ &= 12.0 \, \mu F \end{aligned}$$

Since charge must be conserved,

$$\begin{aligned} V &= \frac{Q_{0}}{C_{eq}} \\ &= 80 \, V \end{aligned}$$

Charge on $C_{1}$:

$$\begin{aligned} Q_{1} &= C_{1} V \\ &= 640 \, \mu C \end{aligned}$$

Charge on $C_{2}$:

$$\begin{aligned} Q_{2} &= C_{2} V \\ &= 320 \, \mu C \end{aligned}$$

Total energy of system:

$$\begin{aligned} U &= \frac{1}{2} Q_{1} V + \frac{1}{2} Q_{2} V \\ &= \frac{1}{2} Q_{0} V \\ &= 0.038 \, J \end{aligned}$$

Notice that the total energy decreased.

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