To solve this problem, let's go step-by-step:
1. **Calculate the capacitance of the initial air-filled capacitor:**
- Plate area \( A = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \)
- Distance between plates \( d = 0.4 \, \text{mm} = 0.4 \times 10^{-3} \, \text{m} \)
- Dielectric constant for air \( \epsilon_r = 1 \)
- Permittivity of free space \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \)
The capacitance \( C_1 \) is given by:
\[
C_1 = \frac{\epsilon_0 \epsilon_r A}{d}
\]
Substituting the values:
\[
C_1 = \frac{8.85 \times 10^{-12} \times 4 \times 10^{-4}}{0.4 \times 10^{-3}}
\]
\[
C_1 = \frac{8.85 \times 10^{-12} \times 4}{0.4 \times 10^{-3}}
\]
\[
C_1 = 88.5 \times 10^{-12} \, \text{F} = 88.5 \, \text{pF}
\]
2. **Initial charge on the air-filled capacitor:**
- Charge \( Q = 25 \, \mu \text{C} = 25 \times 10^{-6} \, \text{C} \)
3. **Calculate the potential difference \( V \) across the air-filled capacitor:**
\[
V = \frac{Q}{C_1}
\]
Substituting the values:
\[
V = \frac{25 \times 10^{-6}}{88.5 \times 10^{-12}}
\]
\[
V = 282.49 \, \text{V}
\]
4. **Calculate the capacitance of the identical capacitor filled with a dielectric medium:**
- Dielectric constant \( \kappa = 8 \)
The capacitance \( C_2 \) is given by:
\[
C_2 = \kappa C_1 = 8 \times 88.5 \, \text{pF} = 708 \, \text{pF}
\]
5. **Calculate the equivalent capacitance when the two capacitors are connected in parallel:**
\[
C_{\text{eq}} = C_1 + C_2 = 88.5 \, \text{pF} + 708 \, \text{pF} = 796.5 \, \text{pF}
\]
6. **Calculate the total charge \( Q_{\text{total}} \):**
- Initial charge on the air-filled capacitor: \( Q = 25 \, \mu \text{C} \)
When connected in parallel, the total charge is the same:
\[
Q_{\text{total}} = 25 \, \mu \text{C}
\]
7. **Calculate the voltage across the combination:**
\[
V_{\text{total}} = \frac{Q_{\text{total}}}{C_{\text{eq}}}
\]
\[
V_{\text{total}} = \frac{25 \times 10^{-6}}{796.5 \times 10^{-12}}
\]
\[
V_{\text{total}} = 31.39 \, \text{V}
\]
8. **Calculate the charge on each capacitor in the new configuration:**
- Charge on \( C_1 \): \( Q_1' = C_1 \times V_{\text{total}} \)
\[
Q_1' = 88.5 \times 10^{-12} \times 31.39 = 2.78 \times 10^{-9} \, \text{C} = 2.78 \, \text{nC}
\]
- Charge on \( C_2 \): \( Q_2' = C_2 \times V_{\text{total}} \)
\[
Q_2' = 708 \times 10^{-12} \times 31.39 = 22.22 \times 10^{-9} \, \text{C} = 22.22 \, \text{nC}
\]
9. **Calculate the charge flow:**
- Initial charge on \( C_1 \): \( 25 \, \mu \text{C} \)
- New charge on \( C_1 \): \( 2.78 \, \text{nC} = 2.78 \times 10^{-3} \, \mu \text{C} \)
- Charge flow \( Q_{\text{flow}} \):
\[
Q_{\text{flow}} = 25 \, \mu \text{C} - 2.78 \times 10^{-3} \, \mu \text{C} = 24.99722 \, \mu \text{C}
\]
Therefore, the amount of charge that flows is approximately \( 24.99722 \, \mu \text{C} \).
