To solve the problem step by step, let's break it down into the required parts:
### Given Data:
- Charge on the first conductor, \( Q_1 = 30 \, \mu C = 30 \times 10^{-6} \, C \)
- Capacitance of the first conductor, \( C_1 = 5 \, \mu F = 5 \times 10^{-6} \, F \)
- Total charge on the second conductor, \( Q_2 = 50 \, \mu C = 50 \times 10^{-6} \, C \)
- Capacitance of the second conductor, \( C_2 = 10 \, \mu F = 10 \times 10^{-6} \, F \)
### (i) Potential of the Conductor
The potential \( V \) of a conductor is given by the formula:
\[
V = \frac{Q}{C}
\]
For the first conductor:
\[
V_1 = \frac{Q_1}{C_1} = \frac{30 \times 10^{-6}}{5 \times 10^{-6}} = 6 \, V
\]
### (ii) Energy Stored in the Electric Field of the Conductor
The energy \( U \) stored in a capacitor is given by:
\[
U = \frac{1}{2} C V^2
\]
Substituting the values:
\[
U_1 = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times 5 \times 10^{-6} \times (6)^2
\]
\[
U_1 = \frac{1}{2} \times 5 \times 10^{-6} \times 36 = \frac{180 \times 10^{-6}}{2} = 90 \times 10^{-6} \, J = 90 \, \mu J
\]
### (iii) Connecting to Another Isolated Conductor
When the first conductor is connected to the second conductor, they will share a common potential \( V_c \).
#### (a) Find the Common Potential of Both Conductors
The total charge after connection is:
\[
Q_{total} = Q_1 + Q_2 = 30 \, \mu C + 50 \, \mu C = 80 \, \mu C
\]
The total capacitance when connected is:
\[
C_{total} = C_1 + C_2 = 5 \, \mu F + 10 \, \mu F = 15 \, \mu F
\]
The common potential \( V_c \) is given by:
\[
V_c = \frac{Q_{total}}{C_{total}} = \frac{80 \times 10^{-6}}{15 \times 10^{-6}} = \frac{80}{15} = \frac{16}{3} \, V \approx 5.33 \, V
\]
#### (b) Find the Heat Dissipated During the Process of Charge Distribution
The heat dissipated \( \Delta H \) is given by the difference in energy before and after connection:
\[
\Delta H = U_{initial} - U_{final}
\]
Where:
- \( U_{initial} = U_1 + U_2 \)
- \( U_2 = \frac{1}{2} C_2 V_2^2 \) where \( V_2 = \frac{Q_2}{C_2} = \frac{50 \times 10^{-6}}{10 \times 10^{-6}} = 5 \, V \)
Calculating \( U_2 \):
\[
U_2 = \frac{1}{2} \times 10 \times 10^{-6} \times (5)^2 = \frac{1}{2} \times 10 \times 10^{-6} \times 25 = 125 \times 10^{-6} \, J = 125 \, \mu J
\]
Now, calculate \( U_{initial} \):
\[
U_{initial} = U_1 + U_2 = 90 \, \mu J + 125 \, \mu J = 215 \, \mu J
\]
Now, calculate \( U_{final} \):
\[
U_{final} = \frac{1}{2} C_{total} V_c^2 = \frac{1}{2} \times 15 \times 10^{-6} \times \left(\frac{16}{3}\right)^2
\]
Calculating \( U_{final} \):
\[
U_{final} = \frac{1}{2} \times 15 \times 10^{-6} \times \frac{256}{9} = \frac{15 \times 256 \times 10^{-6}}{18} = \frac{3840 \times 10^{-6}}{18} \approx 213.33 \, \mu J
\]
Now, calculate the heat dissipated:
\[
\Delta H = 215 \, \mu J - 213.33 \, \mu J \approx 1.67 \, \mu J
\]
#### (c) Find the Ratio of Final Charges on Conductors
Let \( Q_1' \) and \( Q_2' \) be the final charges on the first and second conductors respectively. The final charges can be calculated as:
\[
Q_1' = C_1 V_c = 5 \, \mu F \times \frac{16}{3} \, V = \frac{80}{3} \, \mu C \approx 26.67 \, \mu C
\]
\[
Q_2' = C_2 V_c = 10 \, \mu F \times \frac{16}{3} \, V = \frac{160}{3} \, \mu C \approx 53.33 \, \mu C
\]
The ratio of final charges is:
\[
\text{Ratio} = \frac{Q_1'}{Q_2'} = \frac{26.67}{53.33} = \frac{1}{2}
\]
#### (d) Find the Final Charges on Each Conductor
From the calculations above:
- Final charge on the first conductor \( Q_1' \approx 26.67 \, \mu C \)
- Final charge on the second conductor \( Q_2' \approx 53.33 \, \mu C \)
### Summary of Answers:
1. Potential of the conductor: \( 6 \, V \)
2. Energy stored in the electric field: \( 90 \, \mu J \)
3. Common potential after connection: \( \approx 5.33 \, V \)
4. Heat dissipated: \( \approx 1.67 \, \mu J \)
5. Ratio of final charges: \( 1:2 \)
6. Final charges: \( Q_1' \approx 26.67 \, \mu C \), \( Q_2' \approx 53.33 \, \mu C \)
