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 charges is \(10 \, cm\) (or \(0.1 \, m\)).
- The charges are brought \(4 \, cm\) closer, so the final distance becomes:
\[ r_2 = 10 \, cm - 4 \, cm = 6 \, cm = 0.06 \, m \]

### Step 2: Calculate the Initial Potential Energy
- The potential energy \(U_1\) at the initial distance \(r_1 = 0.1 \, m\) is given by the formula:
\[ U_1 = \frac{k \cdot q_1 \cdot q_2}{r_1} \]
where \(k = 9 \times 10^9 \, N \cdot m^2/C^2\), \(q_1 = 12 \times 10^{-6} \, C\), and \(q_2 = 8 \times 10^{-6} \, C\).

Plugging in the values:
\[ U_1 = \frac{9 \times 10^9 \cdot (12 \times 10^{-6}) \cdot (8 \times 10^{-6})}{0.1} \]

### Step 3: Calculate the Final Potential Energy
- The potential energy \(U_2\) at the final distance \(r_2 = 0.06 \, m\) is:
\[ U_2 = \frac{k \cdot q_1 \cdot q_2}{r_2} \]
Plugging in the values:
\[ U_2 = \frac{9 \times 10^9 \cdot (12 \times 10^{-6}) \cdot (8 \times 10^{-6})}{0.06} \]

### Step 4: Calculate the Work Done
- The work done \(W\) in bringing the charges closer is the change in potential energy:
\[ W = U_2 - U_1 \]
Substituting the expressions for \(U_1\) and \(U_2\):
\[ W = \left(\frac{9 \times 10^9 \cdot (12 \times 10^{-6}) \cdot (8 \times 10^{-6})}{0.06}\right) - \left(\frac{9 \times 10^9 \cdot (12 \times 10^{-6}) \cdot (8 \times 10^{-6})}{0.1}\right) \]

### Step 5: Simplify and Calculate
- 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 in the fractions:
\[ \frac{1}{0.06} - \frac{1}{0.1} = \frac{10 - 6}{0.6} = \frac{4}{0.6} = \frac{20}{3} \]
- Now substitute back to find \(W\):
\[ W = 9 \times 10^9 \cdot (12 \times 10^{-6}) \cdot (8 \times 10^{-6}) \cdot \frac{20}{3} \]

### Step 6: Final Calculation
- Calculate the numerical value:
\[ W = 9 \times 10^9 \cdot 96 \times 10^{-12} \cdot \frac{20}{3} \]
\[ W = 9 \times 20 \times 32 \times 10^{-3} = 576 \, J \]
\[ W \approx 5.76 \, J \]

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