Problem

Two parallel plate capacitors $A$ and $B$, with capacitances of $2 \, \mathrm{\mu F}$ and $6 \, \mathrm{\mu F}$, are charged separately to $120 \, \mathrm{V}$. Then, the positive plate of $A$ is connected to the negative plate of $B$, and the negative plate of $A$ is connected to the positive plate of $B$. Find the final charge on each capacitor.

Data

• Capacitor $C_A = 2 \, \mathrm{\mu F} = 2 \times 10^{-6} \, \mathrm{F}$
• Capacitor $C_B = 6 \, \mathrm{\mu F} = 6 \times 10^{-6} \, \mathrm{F}$
• Voltage $V_A = V_B = 120 \, \mathrm{V}$.

Prerequisite Concepts

• Initial Charge:

$$Q = C \cdot V$$

• Conservation of Charge:

$$Q_{\text{final}} = Q_B - Q_A$$

• Final Voltage:

$$V = \frac{Q_{\text{final}}}{C_A + C_B}$$

Answer

Step 1: Calculate Initial Charges

For capacitor $A$:

$$Q_A = C_A \cdot V_A$$ $$Q_A = 2 \times 10^{-6} \cdot 120 = 240 \, \mathrm{\mu C}$$

For capacitor $B$:

$$Q_B = C_B \cdot V_B$$ $$Q_B = 6 \times 10^{-6} \cdot 120 = 720 \, \mathrm{\mu C}$$

Step 2: Calculate Final Charge

$$Q_{\text{final}} = Q_B - Q_A$$ $$Q_{\text{final}} = 720 - 240 = 480 \, \mathrm{\mu C}$$

Step 3: Calculate Final Voltage

$$V = \frac{Q_{\text{final}}}{C_A + C_B}$$ $$V = \frac{480}{2 + 6} = 60 \, \mathrm{V}$$

Step 4: Calculate Final Charges

For capacitor $A$:

$$Q_A = C_A \cdot V$$ $$Q_A = 2 \cdot 60 = 120 \, \mathrm{\mu C}$$

For capacitor $B$:

$$Q_B = C_B \cdot V$$ $$Q_B = 6 \cdot 60 = 360 \, \mathrm{\mu C}$$

Final Answer

• Final Charge on $A$: $120 \, \mathrm{\mu C}$
• Final Charge on $B$: $360 \, \mathrm{\mu C}$.

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