Let's solve this using the direct formula.

Formula:

The final common voltage

$V$

across the capacitors after they are connected is given by:

$$V = \frac{C_1 V_1 - C_2 V_2}{C_1 + C_2}$$

Step 1: Given Data

• 
  $C_1 = 1 \, \mu\text{F}, \, Q_1 = 1 \, \mu\text{C} \Rightarrow V_1 = \frac{Q_1}{C_1} = \frac{1}{1} = 1 \, \text{V}$ 
• 
  $C_2 = 2 \, \mu\text{F}, \, Q_2 = 4 \, \mu\text{C} \Rightarrow V_2 = \frac{Q_2}{C_2} = \frac{4}{2} = 2 \, \text{V}$ 

Step 2: Final Voltage

Substitute into the formula:

$$V = \frac{C_1 V_1 - C_2 V_2}{C_1 + C_2} = \frac{1 \cdot 1 - 2 \cdot 2}{1 + 2} = \frac{1 - 4}{3} = \frac{-3}{3} = -1 \, \text{V}.$$

(Note: The negative sign indicates the direction of potential difference.)

Step 3: Final Charges

The final charges on the capacitors are calculated using

$Q = C \cdot V$

:

• For
  $C_1$ 

  : $Q_1' = C_1 \cdot V = 1 \cdot (-1) = -1 \, \mu\text{C}$ 

  , but in magnitude, $Q_1' = 1 \, \mu\text{C}$ 

  .

• For
  $C_2$ 

  : $Q_2' = C_2 \cdot V = 2 \cdot (-1) = -2 \, \mu\text{C}$ 

  , but in magnitude, $Q_2' = 2 \, \mu\text{C}$ 

  .

Final Answer:

The final charges are:

$$\boxed{1 \, \mu\text{C} \, \text{and} \, 2 \, \mu\text{C}}.$$