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 find the work needed to bring two positive point charges of 12 µC and 5 µC closer from an initial distance of 10 cm to a final distance of 6 cm, we will calculate the change in potential energy of the system.

### Step-by-Step Solution:

1. **Identify the Charges and Initial Distance**:
- Charge \( q_1 = 12 \, \mu C = 12 \times 10^{-6} \, C \)
- Charge \( q_2 = 5 \, \mu C = 5 \times 10^{-6} \, C \)
- Initial distance \( r_1 = 10 \, cm = 0.1 \, m \)

2. **Calculate the Initial Potential Energy**:
The formula for the potential energy \( U \) between two point charges is given by:
\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \]
where \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \) is Coulomb's constant.

Substituting the values for initial potential energy:
\[ U_{initial} = \frac{9 \times 10^9 \cdot (12 \times 10^{-6}) \cdot (5 \times 10^{-6})}{0.1} \]
\[ U_{initial} = \frac{9 \times 10^9 \cdot 60 \times 10^{-12}}{0.1} \]
\[ U_{initial} = \frac{540 \times 10^{-3}}{0.1} = 5.4 \, J \]

3. **Identify the Final Distance**:
- Final distance \( r_2 = 10 \, cm - 4 \, cm = 6 \, cm = 0.06 \, m \)

4. **Calculate the Final Potential Energy**:
Using the same formula for potential energy:
\[ U_{final} = \frac{9 \times 10^9 \cdot (12 \times 10^{-6}) \cdot (5 \times 10^{-6})}{0.06} \]
\[ U_{final} = \frac{9 \times 10^9 \cdot 60 \times 10^{-12}}{0.06} \]
\[ U_{final} = \frac{540 \times 10^{-3}}{0.06} = 9 \, J \]

5. **Calculate the Work Done**:
The work done by the external agent is equal to the change in potential energy:
\[ W = U_{final} - U_{initial} \]
\[ W = 9 \, J - 5.4 \, J = 3.6 \, J \]

### Final Answer:
The work needed to bring the charges 4 cm closer is **3.6 Joules**.