A charged spherical conductor has a surface charge density of `0.7C//m^(2)` .When its charge is increased by `0.44C` the charge denstiy changes by `0.14C//m^(2)` the radius of the sphere is

Options

A. `5cm`
B. `10m`
C. `0.5m`
D. `5m`

Explanation

To find the radius of the charged spherical conductor, we can follow these steps:

### Step 1: Understand the given data
We have:
- Initial surface charge density, \( \sigma_1 = 0.7 \, \text{C/m}^2 \)
- Increase in charge, \( \Delta Q = 0.44 \, \text{C} \)
- Change in surface charge density, \( \Delta \sigma = 0.14 \, \text{C/m}^2 \)

### Step 2: Calculate the new surface charge density
The new surface charge density after the increase in charge is:
\[ \sigma_2 = \sigma_1 + \Delta \sigma = 0.7 \, \text{C/m}^2 + 0.14 \, \text{C/m}^2 = 0.84 \, \text{C/m}^2 \]

### Step 3: Relate surface charge density to total charge
The surface charge density is related to the total charge \( Q \) and the radius \( R \) of the sphere by the formula:
\[ \sigma = \frac{Q}{A} \]
where \( A \) is the surface area of the sphere, given by \( A = 4\pi R^2 \). Thus, we can express the total charge as:
\[ Q = \sigma \cdot 4\pi R^2 \]

### Step 4: Write equations for the initial and final charges
1. For the initial charge:
\[ Q_1 = \sigma_1 \cdot 4\pi R^2 = 0.7 \cdot 4\pi R^2 \]
2. For the final charge after the increase:
\[ Q_2 = \sigma_2 \cdot 4\pi R^2 = 0.84 \cdot 4\pi R^2 \]

### Step 5: Set up the equation for the increase in charge
The increase in charge can be expressed as:
\[ Q_2 = Q_1 + \Delta Q \]
Substituting the expressions for \( Q_1 \) and \( Q_2 \):
\[ 0.84 \cdot 4\pi R^2 = 0.7 \cdot 4\pi R^2 + 0.44 \]

### Step 6: Simplify the equation
We can factor out \( 4\pi R^2 \) from both sides:
\[ 0.84 = 0.7 + \frac{0.44}{4\pi R^2} \]
Subtract \( 0.7 \) from both sides:
\[ 0.14 = \frac{0.44}{4\pi R^2} \]

### Step 7: Solve for \( R^2 \)
Rearranging gives:
\[ 4\pi R^2 = \frac{0.44}{0.14} \]
Calculating the right side:
\[ 4\pi R^2 = \frac{0.44}{0.14} \approx 3.142857 \]
Now, divide both sides by \( 4\pi \):
\[ R^2 = \frac{3.142857}{4\pi} \]
Calculating \( R^2 \):
\[ R^2 \approx \frac{3.142857}{12.56637} \approx 0.25 \]
Taking the square root gives:
\[ R \approx 0.5 \, \text{m} \]

### Final Answer
The radius of the sphere is approximately \( R \approx 0.5 \, \text{m} \).

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