Two ball bearings, one having a radius of 2 mm and a surface charge density equal to `5muC//m^(2)` and another having a radius of 1mm and a surface density of charge equal to `-2muC//m^(2)` are situated inside a closed surface. The T.N.E.I. over the closed surface is

Options

A. `1.262xx10^(-10)C`
B. `1.262xx10^(-8)C`
C. `2.262xx10^(-8)C`
D. `2.262xx10^(-10)C`

Explanation

To solve the problem, we need to calculate the Total Normal Electric Induction (TNEI) over a closed surface containing two ball bearings with given surface charge densities and radii.

### Step-by-Step Solution:

1. **Understand the Given Data**:
- For the first ball bearing:
- Radius \( R_1 = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- Surface charge density \( \sigma_1 = 5 \, \mu\text{C/m}^2 = 5 \times 10^{-6} \, \text{C/m}^2 \)
- For the second ball bearing:
- Radius \( R_2 = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Surface charge density \( \sigma_2 = -2 \, \mu\text{C/m}^2 = -2 \times 10^{-6} \, \text{C/m}^2 \)

2. **Calculate the Charge on Each Ball Bearing**:
- The charge \( Q \) on a spherical object can be calculated using the formula:
\[ Q = \sigma \times \text{Surface Area} \]
- The surface area \( A \) of a sphere is given by:
\[ A = 4 \pi R^2 \]
- For the first ball bearing:
\[ Q_1 = \sigma_1 \times 4 \pi R_1^2 = 5 \times 10^{-6} \times 4 \pi (2 \times 10^{-3})^2 \]
- For the second ball bearing:
\[ Q_2 = \sigma_2 \times 4 \pi R_2^2 = -2 \times 10^{-6} \times 4 \pi (1 \times 10^{-3})^2 \]

3. **Calculate Each Charge**:
- Calculate \( Q_1 \):
\[ Q_1 = 5 \times 10^{-6} \times 4 \pi (4 \times 10^{-6}) = 5 \times 10^{-6} \times 16 \pi \times 10^{-6} \]
\[ Q_1 = 80 \pi \times 10^{-12} \, \text{C} \]
- Calculate \( Q_2 \):
\[ Q_2 = -2 \times 10^{-6} \times 4 \pi (1 \times 10^{-6}) = -2 \times 10^{-6} \times 4 \pi \times 10^{-6} \]
\[ Q_2 = -8 \pi \times 10^{-12} \, \text{C} \]

4. **Calculate Total Charge Enclosed**:
- The total charge \( Q_{\text{total}} \) is given by:
\[ Q_{\text{total}} = Q_1 + Q_2 \]
\[ Q_{\text{total}} = (80 \pi - 8 \pi) \times 10^{-12} = 72 \pi \times 10^{-12} \, \text{C} \]

5. **Calculate TNEI**:
- The Total Normal Electric Induction (TNEI) is proportional to the total charge enclosed:
\[ \text{TNEI} = \frac{Q_{\text{total}}}{\epsilon_0} \]
- However, since we are only asked for the total charge contribution, we can express TNEI directly as:
\[ \text{TNEI} = 72 \pi \times 10^{-12} \, \text{C} \]
- Approximating \( \pi \approx 3.14 \):
\[ \text{TNEI} \approx 72 \times 3.14 \times 10^{-12} \approx 226.08 \times 10^{-12} \, \text{C} \]

### Final Answer:
\[ \text{TNEI} \approx 226.08 \, \text{pC} \, (\text{picoCoulombs}) \]