Problem

A $6 \, \mathrm{\mu F}$ capacitor is charged to a potential difference of $120 \, \mathrm{V}$ and then connected to an uncharged $4 \, \mathrm{\mu F}$ capacitor. Calculate the potential difference across the capacitors after they are connected.

Data

• Capacitor $C_1 = 6 \, \mathrm{\mu F}$
• Capacitor $C_2 = 4 \, \mathrm{\mu F}$
• Initial Voltage $V_1 = 120 \, \mathrm{V}$.

Prerequisite Concepts

• Initial Charge on Charged Capacitor:

$$Q_1 = C_1 \cdot V_1$$

• Charge Conservation:

$$Q_{\text{total}} = Q_1 + Q_2$$

• Final Voltage Across Both Capacitors:

$$V = \frac{Q_{\text{total}}}{C_1 + C_2}$$

Answer

Step 1: Calculate Initial Charge on $C_1$

$$Q_1 = C_1 \cdot V_1$$ $$Q_1 = 6 \cdot 120 = 720 \, \mathrm{\mu C}$$

Step 2: Total Charge in the System

Initially, $C_2$ is uncharged, so the total charge is:

$$Q_{\text{total}} = Q_1 = 720 \, \mathrm{\mu C}$$

Step 3: Total Capacitance After Connection

Since the capacitors are connected in parallel:

$$C_{\text{total}} = C_1 + C_2$$ $$C_{\text{total}} = 6 + 4 = 10 \, \mathrm{\mu F}$$

Step 4: Final Voltage

$$V = \frac{Q_{\text{total}}}{C_{\text{total}}}$$ $$V = \frac{720}{10} = 72 \, \mathrm{V}$$

Final Answer

The potential difference across both capacitors after connection is $72 \, \mathrm{V}$.

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