When `30mu C` charge is given to an isolated conductor of capacitance `5mu F`. Find out the following
(i) Potential of the conductor
(ii) Energy stored in the electric field of conductor
(iii) If this conductor is now connected to another isolated conductor by a conducting wire (at very large distance ) of total charge `50mu C` and capacity `10mu F` then
(a) find out the common potential of both the conductors.
(b) Find out the heat dissipated during the process of charge distribution.
(c) Find out the ratio of final charges on conductors.
d) Find out the final charges on each conductor.

Answer

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: \( Q_1' \approx 26.67 \, \mu C \), \( Q_2' \approx 53.33 \, \mu C \)