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 J`
C. `13 J`
D. `13 eV`

Explanation

To solve the problem of finding the work done in bringing two positive point charges of `12 µC` and `8 µC` closer by `4 cm`, we can follow these steps:

### Step 1: Understand the Initial and Final Distances
- The initial distance between the two charges (R1) is `10 cm` or `0.1 m`.
- The final distance after bringing them `4 cm` closer (R2) is `10 cm - 4 cm = 6 cm` or `0.06 m`.

### Step 2: Write the Formula for Work Done
The work done in moving the charges is equal to the change in potential energy (ΔPE), which can be calculated using the formula:
\[ \Delta PE = PE_{final} - PE_{initial} \]
Where:
- \( PE = k \frac{Q_1 Q_2}{R} \)
- \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).
- \( Q_1 = 12 \, \mu C = 12 \times 10^{-6} \, C \)
- \( Q_2 = 8 \, \mu C = 8 \times 10^{-6} \, C \)

### Step 3: Calculate the Initial Potential Energy
Using the initial distance (R1 = 0.1 m):
\[ PE_{initial} = k \frac{Q_1 Q_2}{R_1} = 9 \times 10^9 \frac{(12 \times 10^{-6})(8 \times 10^{-6})}{0.1} \]
\[ PE_{initial} = 9 \times 10^9 \frac{96 \times 10^{-12}}{0.1} = 9 \times 10^9 \times 9.6 \times 10^{-10} = 8.64 \times 10^{-1} \, J \]

### Step 4: Calculate the Final Potential Energy
Using the final distance (R2 = 0.06 m):
\[ PE_{final} = k \frac{Q_1 Q_2}{R_2} = 9 \times 10^9 \frac{(12 \times 10^{-6})(8 \times 10^{-6})}{0.06} \]
\[ PE_{final} = 9 \times 10^9 \frac{96 \times 10^{-12}}{0.06} = 9 \times 10^9 \times 1.6 \times 10^{-9} = 14.4 \, J \]

### Step 5: Calculate the Change in Potential Energy
Now, we can find the change in potential energy:
\[ \Delta PE = PE_{final} - PE_{initial} = 14.4 - 0.864 = 13.536 \, J \]

### Step 6: Conclusion
The work done in bringing the two charges `4 cm` closer is approximately:
\[ W = \Delta PE = 13.536 \, J \]