To solve the problem step by step, we will follow the principles of capacitors, charge conservation, and voltage relationships.
### Step 1: Calculate the initial charge on the 6 µF capacitor
The charge \( Q_1 \) on the 6 µF capacitor when it is charged to 100 V can be calculated using the formula:
\[
Q = C \cdot V
\]
Substituting the values:
\[
Q_1 = 6 \times 10^{-6} \, \text{F} \times 100 \, \text{V} = 600 \times 10^{-6} \, \text{C} = 600 \, \mu C
\]
### Step 2: Set up the equations for the connected capacitors
When the charged capacitor is connected to the uncharged 14 µF capacitor, the total charge is conserved. Let \( Q_2 \) be the charge on the 14 µF capacitor after they are connected. The total charge before they are connected is:
\[
Q_{\text{total}} = Q_1 + Q_2
\]
Since the 14 µF capacitor is initially uncharged, we have:
\[
Q_{\text{total}} = Q_1 = 600 \, \mu C
\]
### Step 3: Relate the charges and voltages after connection
After connecting the two capacitors, they will reach a common voltage \( V_c \). The relationship between charge, capacitance, and voltage gives us:
\[
Q_1 = C_1 \cdot V_c \quad \text{and} \quad Q_2 = C_2 \cdot V_c
\]
This leads to:
\[
Q_1 = 6 \times 10^{-6} \cdot V_c \quad \text{and} \quad Q_2 = 14 \times 10^{-6} \cdot V_c
\]
### Step 4: Write the conservation of charge equation
From charge conservation, we have:
\[
Q_1 + Q_2 = 600 \times 10^{-6}
\]
Substituting the expressions for \( Q_1 \) and \( Q_2 \):
\[
6 \times 10^{-6} \cdot V_c + 14 \times 10^{-6} \cdot V_c = 600 \times 10^{-6}
\]
Factoring out \( V_c \):
\[
(6 + 14) \times 10^{-6} \cdot V_c = 600 \times 10^{-6}
\]
\[
20 \times 10^{-6} \cdot V_c = 600 \times 10^{-6}
\]
Solving for \( V_c \):
\[
V_c = \frac{600 \times 10^{-6}}{20 \times 10^{-6}} = 30 \, \text{V}
\]
### Step 5: Calculate the charges on each capacitor
Now we can find \( Q_1 \) and \( Q_2 \) using the common voltage:
\[
Q_1 = 6 \times 10^{-6} \cdot 30 = 180 \times 10^{-6} \, \text{C} = 180 \, \mu C
\]
\[
Q_2 = 14 \times 10^{-6} \cdot 30 = 420 \times 10^{-6} \, \text{C} = 420 \, \mu C
\]
### Step 6: Find the ratio of the charges
The ratio of the charges on the 6 µF and 14 µF capacitors is:
\[
\text{Ratio} = \frac{Q_1}{Q_2} = \frac{180 \, \mu C}{420 \, \mu C} = \frac{180}{420} = \frac{3}{7}
\]
### Final Results
- The ratio of the charges \( Q_1 : Q_2 = 3 : 7 \)
- The potential of the 6 µF capacitor \( V_1 = 30 \, \text{V} \)
