A metallic sphere of radius 18 cm has been given a charge of `5xx10^(-6)C.` The energy of the charged conductor is

Options

A. `0.2J`
B. `0.6J`
C. `1.2J`
D. `2.4J`

Explanation

To find the energy of a charged metallic sphere, we can follow these steps:

### Step 1: Identify the given values
- Radius of the sphere, \( R = 18 \, \text{cm} = 0.18 \, \text{m} \)
- Charge on the sphere, \( Q = 5 \times 10^{-6} \, \text{C} \)

### Step 2: Calculate the capacitance of the sphere
The formula for the capacitance \( C \) of a metallic sphere is given by:
\[ C = 4 \pi \epsilon_0 R \]
where \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, \text{F/m} \).

Substituting the values:
\[ C = 4 \pi (8.85 \times 10^{-12}) (0.18) \]
Calculating \( C \):
\[ C \approx 4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.18 \]
\[ C \approx 2.00 \times 10^{-12} \, \text{F} \]

### Step 3: Calculate the energy stored in the sphere
The energy \( U \) stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
We also know that \( V \) (the potential) can be expressed in terms of charge and capacitance:
\[ V = \frac{Q}{C} \]
Substituting this into the energy formula gives:
\[ U = \frac{1}{2} C \left(\frac{Q}{C}\right)^2 = \frac{1}{2} \frac{Q^2}{C} \]

### Step 4: Substitute the values into the energy formula
Now substituting \( Q = 5 \times 10^{-6} \, \text{C} \) and \( C \approx 2.00 \times 10^{-12} \, \text{F} \):
\[ U = \frac{1}{2} \frac{(5 \times 10^{-6})^2}{2.00 \times 10^{-12}} \]
Calculating \( U \):
\[ U = \frac{1}{2} \frac{25 \times 10^{-12}}{2.00 \times 10^{-12}} = \frac{1}{2} \times 12.5 = 6.25 \, \text{J} \]

### Final Answer
The energy of the charged conductor is approximately \( 6.25 \, \text{J} \).

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