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

Answer

To solve the problem step by step, we will calculate the potential of the conductor and the energy stored in the electric field of the conductor.

### Given:
- Charge, \( Q = 30 \, \mu C = 30 \times 10^{-6} \, C \)
- Capacitance, \( C = 5 \, \mu F = 5 \times 10^{-6} \, F \)

### (i) Finding the Potential of the Conductor

The formula for the potential \( V \) of a conductor is given by:

\[ V = \frac{Q}{C} \]

Substituting the values:

\[ V = \frac{30 \times 10^{-6} \, C}{5 \times 10^{-6} \, F} \]

Calculating this:

\[ V = \frac{30}{5} = 6 \, V \]

Thus, the potential of the conductor is **6 volts**.

### (ii) Finding the Energy Stored in the Electric Field of the Conductor

The energy \( U \) stored in the electric field of a capacitor is given by the formula:

\[ U = \frac{1}{2} \frac{Q^2}{C} \]

Substituting the values:

\[ U = \frac{1}{2} \cdot \frac{(30 \times 10^{-6})^2}{5 \times 10^{-6}} \]

Calculating \( (30 \times 10^{-6})^2 \):

\[ (30 \times 10^{-6})^2 = 900 \times 10^{-12} \, C^2 \]

Now substituting this back into the energy formula:

\[ U = \frac{1}{2} \cdot \frac{900 \times 10^{-12}}{5 \times 10^{-6}} \]

Calculating the division:

\[ \frac{900 \times 10^{-12}}{5 \times 10^{-6}} = 180 \times 10^{-6} \, J \]

Now, multiplying by \( \frac{1}{2} \):

\[ U = \frac{1}{2} \cdot 180 \times 10^{-6} = 90 \times 10^{-6} \, J \]

Thus, the energy stored in the electric field of the conductor is **90 microjoules**.

### Summary of Results:
- Potential of the conductor: **6 volts**
- Energy stored in the electric field: **90 microjoules**