Two positive point charges of `12 mu C` and `8 mu C` are `10 cm` apart. The work done in bringing then `4 cm` closer is

Options

A. 5.8 J
B. 5.8 eV
C. 13 J
D. 13 eV

Answer

To solve the problem of calculating the work done in bringing two positive point charges of \(12 \, \mu C\) and \(8 \, \mu C\) closer by \(4 \, cm\), we can follow these steps:

### Step 1: Understand the Initial and Final Distances
- The initial distance between the two charges is \(10 \, cm\).
- We want to bring them \(4 \, cm\) closer, so the final distance will be:
\[ \text{Final Distance} = 10 \, cm - 4 \, cm = 6 \, cm \]

### Step 2: Convert Distances to Meters
- Convert the distances from centimeters to meters:
\[ r_{\text{initial}} = 10 \, cm = 0.1 \, m \]
\[ r_{\text{final}} = 6 \, cm = 0.06 \, m \]

### Step 3: Calculate the Initial and Final Potential Energies
- The potential energy \(U\) between two point charges is given by the formula:
\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \]
where \(k\) is Coulomb's constant, \(k = 9 \times 10^9 \, N \cdot m^2/C^2\), \(q_1 = 12 \, \mu C = 12 \times 10^{-6} \, C\), and \(q_2 = 8 \, \mu C = 8 \times 10^{-6} \, C\).

- Calculate the initial potential energy \(U_{\text{initial}}\):
\[ U_{\text{initial}} = \frac{9 \times 10^9 \cdot (12 \times 10^{-6}) \cdot (8 \times 10^{-6})}{0.1} \]

- Calculate the final potential energy \(U_{\text{final}}\):
\[ U_{\text{final}} = \frac{9 \times 10^9 \cdot (12 \times 10^{-6}) \cdot (8 \times 10^{-6})}{0.06} \]

### Step 4: Calculate the Work Done
- The work done \(W\) by the external agent is the change in potential energy:
\[ W = U_{\text{final}} - U_{\text{initial}} \]

- Substitute the expressions for potential energies:
\[ W = \left( \frac{9 \times 10^9 \cdot (12 \times 10^{-6}) \cdot (8 \times 10^{-6})}{0.06} \right) - \left( \frac{9 \times 10^9 \cdot (12 \times 10^{-6}) \cdot (8 \times 10^{-6})}{0.1} \right) \]

### Step 5: Simplify the Expression
- Factor out the common terms:
\[ W = 9 \times 10^9 \cdot (12 \times 10^{-6}) \cdot (8 \times 10^{-6}) \left( \frac{1}{0.06} - \frac{1}{0.1} \right) \]

- Calculate the difference:
\[ \frac{1}{0.06} - \frac{1}{0.1} = \frac{100}{6} - 10 = \frac{100 - 60}{6} = \frac{40}{6} \approx 6.67 \]

### Step 6: Final Calculation
- Substitute back to find \(W\):
\[ W = 9 \times 10^9 \cdot (12 \times 10^{-6}) \cdot (8 \times 10^{-6}) \cdot \frac{40}{6} \]
- Calculate the numerical value:
\[ W = 9 \times 10^9 \cdot 96 \times 10^{-12} \cdot \frac{40}{6} \]
\[ W = 864 \times 10^{-3} \cdot \frac{40}{6} \]
\[ W \approx 864 \times 10^{-3} \cdot 6.67 \approx 5.76 \, J \]

### Final Answer
The work done in bringing the charges \(4 \, cm\) closer is approximately \(5.76 \, J\).