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. 3.6 J
B. 2.6 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 closer together, we can follow these steps:

### Step-by-Step Solution:

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

2. **Calculate the Initial Potential Energy (\( U_i \))**:
- The formula for the potential energy between two point charges is given by:
\[ U_i = k \frac{q_1 q_2}{r_1} \]
- Where \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \) (Coulomb's constant).
- Substituting the values:
\[ U_i = 9 \times 10^9 \times \frac{(12 \times 10^{-6})(5 \times 10^{-6})}{0.1} \]
- Simplifying:
\[ U_i = 9 \times 10^9 \times \frac{60 \times 10^{-12}}{0.1} = 9 \times 10^9 \times 600 \times 10^{-12} = 5400 \times 10^{-3} = 5.4 \, J \]

3. **Calculate the Final Distance**:
- The charges are brought 4 cm closer, so the new distance \( r_2 \) is:
\[ r_2 = 10 \, cm - 4 \, cm = 6 \, cm = 0.06 \, m \]

4. **Calculate the Final Potential Energy (\( U_f \))**:
- Using the same formula for potential energy:
\[ U_f = k \frac{q_1 q_2}{r_2} \]
- Substituting the new distance:
\[ U_f = 9 \times 10^9 \times \frac{(12 \times 10^{-6})(5 \times 10^{-6})}{0.06} \]
- Simplifying:
\[ U_f = 9 \times 10^9 \times \frac{60 \times 10^{-12}}{0.06} = 9 \times 10^9 \times 1000 \times 10^{-12} = 9000 \times 10^{-3} = 9 \, J \]

5. **Calculate the Work Done (\( W \))**:
- The work done by the external agent is the change in potential energy:
\[ W = U_f - U_i \]
- Substituting the values:
\[ W = 9 \, J - 5.4 \, J = 3.6 \, J \]

### Final Answer:
The work needed to bring the two charges 4 cm closer is \( 3.6 \, J \).

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