Two point charges `A=+3nC and B = +1nC` are placed 5 cm apart in the air. The work done to move charge B towards A by 1 cm is

Options

A. `2.0xx10^(-7)J`
B. `2.7xx10^(-7)J`
C. `12.1xx10^(-7)J`
D. `1.35xx10^(-7)J`

Answer

To solve the problem of finding the work done to move charge B towards charge A by 1 cm, we can follow these steps:

### Step 1: Understand the Initial Setup
- We have two point charges:
- Charge A (q1) = +3 nC = 3 × 10^(-9) C
- Charge B (q2) = +1 nC = 1 × 10^(-9) C
- The initial distance (r1) between the charges is 5 cm = 0.05 m.

### Step 2: Calculate the Initial Potential Energy (U_initial)
The formula for the potential energy (U) between two point charges is given by:
\[ U = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 \cdot q_2}{r} \]
Where:
- \( \epsilon_0 \) (the permittivity of free space) = \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)

Substituting the values for initial potential energy:
\[ U_{\text{initial}} = \frac{1}{4 \pi (8.85 \times 10^{-12})} \cdot \frac{(3 \times 10^{-9}) \cdot (1 \times 10^{-9})}{0.05} \]

### Step 3: Calculate the Final Distance
When charge B is moved 1 cm closer to charge A, the new distance (r2) becomes:
\[ r_2 = 5 \, \text{cm} - 1 \, \text{cm} = 4 \, \text{cm} = 0.04 \, \text{m} \]

### Step 4: Calculate the Final Potential Energy (U_final)
Now, we calculate the final potential energy using the new distance:
\[ U_{\text{final}} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 \cdot q_2}{r_2} \]
Substituting the values:
\[ U_{\text{final}} = \frac{1}{4 \pi (8.85 \times 10^{-12})} \cdot \frac{(3 \times 10^{-9}) \cdot (1 \times 10^{-9})}{0.04} \]

### Step 5: Calculate the Work Done (W)
The work done to move charge B towards A is equal to the change in potential energy:
\[ W = U_{\text{final}} - U_{\text{initial}} \]

### Step 6: Substitute and Simplify
1. Calculate \( U_{\text{initial}} \):
\[ U_{\text{initial}} = \frac{1}{4 \pi (8.85 \times 10^{-12})} \cdot \frac{(3 \times 10^{-9}) \cdot (1 \times 10^{-9})}{0.05} \]
\[ U_{\text{initial}} = \frac{9 \times 10^9 \cdot 3 \times 10^{-18}}{0.05} = \frac{27 \times 10^{-9}}{0.05} = 5.4 \times 10^{-7} \, \text{J} \]

2. Calculate \( U_{\text{final}} \):
\[ U_{\text{final}} = \frac{1}{4 \pi (8.85 \times 10^{-12})} \cdot \frac{(3 \times 10^{-9}) \cdot (1 \times 10^{-9})}{0.04} \]
\[ U_{\text{final}} = \frac{9 \times 10^9 \cdot 3 \times 10^{-18}}{0.04} = \frac{27 \times 10^{-9}}{0.04} = 6.75 \times 10^{-7} \, \text{J} \]

3. Now calculate the work done:
\[ W = 6.75 \times 10^{-7} - 5.4 \times 10^{-7} = 1.35 \times 10^{-7} \, \text{J} \]

### Final Answer
The work done to move charge B towards A by 1 cm is:
\[ W = 1.35 \times 10^{-7} \, \text{J} \]