Two positive point charges of 12 and 5 microcoulombs, are placed 10 cm apart in air. The work needed to bring them 4 cm closer is

Options

A. 2.4 J
B. 3.6 J
C. 4.8 J
D. 6.0 J

Answer

To solve the problem of finding the work needed to bring two positive point charges of 12 µC and 5 µC closer by 4 cm, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial and Final Distances:**
- The initial distance between the charges is 10 cm (0.1 m).
- The final distance after bringing them 4 cm closer is 6 cm (0.06 m).

2. **Use the Formula for Work Done:**
- The work done (W) in bringing the charges closer can be calculated using the change in potential energy (U) of the system:
\[ W = U_{\text{final}} - U_{\text{initial}} \]
- The potential energy (U) between two point charges is given by:
\[ U = k \frac{q_1 q_2}{r} \]
- Where \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.

3. **Calculate Initial Potential Energy (U_initial):**
- For the initial distance (0.1 m):
\[ U_{\text{initial}} = k \frac{q_1 q_2}{r_{\text{initial}}} = 8.99 \times 10^9 \frac{(12 \times 10^{-6})(5 \times 10^{-6})}{0.1} \]
- Simplifying this:
\[ U_{\text{initial}} = 8.99 \times 10^9 \frac{60 \times 10^{-12}}{0.1} = 8.99 \times 10^9 \times 600 \times 10^{-12} = 5394 \times 10^{-3} \, \text{J} = 5.394 \, \text{J} \]

4. **Calculate Final Potential Energy (U_final):**
- For the final distance (0.06 m):
\[ U_{\text{final}} = k \frac{q_1 q_2}{r_{\text{final}}} = 8.99 \times 10^9 \frac{(12 \times 10^{-6})(5 \times 10^{-6})}{0.06} \]
- Simplifying this:
\[ U_{\text{final}} = 8.99 \times 10^9 \frac{60 \times 10^{-12}}{0.06} = 8.99 \times 10^9 \times 1000 \times 10^{-12} = 8.99 \times 10^{-3} \, \text{J} = 8.99 \, \text{J} \]

5. **Calculate Work Done (W):**
- Now, substitute the values into the work done equation:
\[ W = U_{\text{final}} - U_{\text{initial}} = 8.99 - 5.394 = 3.596 \, \text{J} \]

### Final Answer:
The work needed to bring the two charges 4 cm closer is approximately **3.596 J**.