A charged spherical conductor has a surface density of `0.07 C cm^(-2)`. When the charge is increased by `4.4 C`, the surface density changes to `0.084 C cm^(-2). Find the initial charge and capacitance of the spherical conductor.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between charge, surface charge density, and area.
The surface charge density \( \sigma \) is defined as the charge \( Q \) per unit area \( A \) of the conductor. For a spherical conductor, the area \( A \) is given by:
\[ A = 4\pi r^2 \]
Thus, the surface charge density can be expressed as:
\[ \sigma = \frac{Q}{4\pi r^2} \]

### Step 2: Set up the equations based on the given information.
1. Initially, the surface charge density is \( \sigma_1 = 0.07 \, \text{C/cm}^2 \):
\[ 0.07 = \frac{Q}{4\pi r^2} \quad \text{(Equation 1)} \]
2. After adding \( 4.4 \, \text{C} \), the new surface charge density is \( \sigma_2 = 0.084 \, \text{C/cm}^2 \):
\[ 0.084 = \frac{Q + 4.4}{4\pi r^2} \quad \text{(Equation 2)} \]

### Step 3: Divide Equation 1 by Equation 2.
To eliminate \( r^2 \), we can divide Equation 1 by Equation 2:
\[ \frac{0.07}{0.084} = \frac{Q}{Q + 4.4} \]
This simplifies to:
\[ \frac{7}{8.4} = \frac{Q}{Q + 4.4} \]
Cross-multiplying gives:
\[ 7(Q + 4.4) = 8.4Q \]

### Step 4: Solve for \( Q \).
Expanding the equation:
\[ 7Q + 30.8 = 8.4Q \]
Rearranging gives:
\[ 30.8 = 8.4Q - 7Q \]
\[ 30.8 = 1.4Q \]
Thus, solving for \( Q \):
\[ Q = \frac{30.8}{1.4} = 22 \, \text{C} \]

### Step 5: Find the radius \( r \) using Equation 1.
Substituting \( Q = 22 \, \text{C} \) back into Equation 1:
\[ 0.07 = \frac{22}{4\pi r^2} \]
Rearranging gives:
\[ r^2 = \frac{22}{4\pi \times 0.07} \]
Calculating \( r^2 \):
\[ r^2 = \frac{22}{0.28\pi} \approx \frac{22}{0.8796} \approx 25 \, \text{m}^2 \]
Taking the square root:
\[ r \approx 5 \, \text{m} \]

### Step 6: Calculate the capacitance \( C \).
The capacitance \( C \) of a spherical conductor is given by:
\[ C = 4\pi \epsilon_0 r \]
Using \( \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m} \) and \( r = 5 \, \text{m} \):
\[ C = 4\pi (8.85 \times 10^{-12}) (5) \]
Calculating:
\[ C \approx 4 \times 3.14 \times 8.85 \times 10^{-12} \times 5 \]
\[ C \approx 5.56 \times 10^{-12} \, \text{F} \]

### Final Answers:
- Initial charge \( Q = 22 \, \text{C} \)
- Capacitance \( C \approx 5.56 \times 10^{-12} \, \text{F} \)