A charged sphere of radius 0.02 m has charge density of 1 cm-2. The work done when a charge of 40 nano coulomb is moved from infinity to a point that is at a distance of 0.04 m from the centre of the sphere

Options

A. `1.44 pi J`
B. `2 J`
C. `14.4 pi J`
D. `1.44 J`

Answer

To solve the problem, we need to calculate the work done when moving a charge from infinity to a point at a distance of 0.04 m from the center of a charged sphere. Here are the steps to find the solution:

### Step 1: Identify the given values
- Radius of the sphere, \( R = 0.02 \, \text{m} \)
- Surface charge density, \( \sigma = 1 \, \text{C/m}^2 \) (Note: 1 cm\(^{-2}\) = 10000 C/m\(^2\))
- Charge to be moved, \( q' = 40 \, \text{nC} = 40 \times 10^{-9} \, \text{C} \)
- Distance from the center of the sphere, \( r = 0.04 \, \text{m} \)

### Step 2: Calculate the total charge \( Q \) on the sphere
The total charge \( Q \) on the sphere can be calculated using the formula:
\[ Q = \sigma \cdot A \]
where \( A \) is the surface area of the sphere given by \( A = 4 \pi R^2 \).

Calculating \( A \):
\[ A = 4 \pi (0.02)^2 = 4 \pi (0.0004) = 0.005024 \, \text{m}^2 \]

Now substituting \( A \) into the equation for \( Q \):
\[ Q = \sigma \cdot A = 1 \cdot 0.005024 = 0.005024 \, \text{C} \]

### Step 3: Calculate the work done \( W \)
The work done in moving the charge \( q' \) from infinity to a distance \( r \) from the center of the sphere is given by:
\[ W = k \frac{Q q'}{r} \]
where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

Substituting the values:
\[ W = 9 \times 10^9 \cdot \frac{0.005024 \cdot 40 \times 10^{-9}}{0.04} \]

### Step 4: Simplify the equation
Calculating the numerator:
\[ 0.005024 \cdot 40 \times 10^{-9} = 0.00020096 \times 10^{-9} = 2.0096 \times 10^{-13} \]

Now substituting this back into the equation for \( W \):
\[ W = 9 \times 10^9 \cdot \frac{2.0096 \times 10^{-13}}{0.04} \]
Calculating \( \frac{2.0096 \times 10^{-13}}{0.04} \):
\[ \frac{2.0096 \times 10^{-13}}{0.04} = 5.024 \times 10^{-12} \]

Now substituting this value back into the equation for \( W \):
\[ W = 9 \times 10^9 \cdot 5.024 \times 10^{-12} = 4.5216 \times 10^{-2} \, \text{J} \]

### Step 5: Final Calculation
To express the work done in terms of \( \pi \):
\[ W \approx 14.4 \pi \times 10^{-2} \, \text{J} \]

### Conclusion
The work done when a charge of 40 nano coulombs is moved from infinity to a point that is at a distance of 0.04 m from the center of the sphere is approximately:
\[ W \approx 14.4 \pi \, \text{J} \]