Two charges `5mu C and 4mu C` are separated by a distanc 20 cm in air. Work to be done to decrease the distance to 10 cm is

Options

A. 1.8 j
B. 0.45 j
C. 2.7 j
D. 0.9 j

Answer

To solve the problem of calculating the work done to decrease the distance between two charges from 20 cm to 10 cm, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Distances**:
- Let \( Q_1 = 5 \, \mu C = 5 \times 10^{-6} \, C \)
- Let \( Q_2 = 4 \, \mu C = 4 \times 10^{-6} \, C \)
- Initial distance \( r_1 = 20 \, cm = 20 \times 10^{-2} \, m = 0.2 \, m \)
- Final distance \( r_2 = 10 \, cm = 10 \times 10^{-2} \, m = 0.1 \, m \)

2. **Use the Formula for Potential Energy**:
- The potential energy \( U \) between two point charges is given by:
\[ U = k \frac{Q_1 Q_2}{r} \]
- Where \( k = 9 \times 10^9 \, N m^2/C^2 \) is Coulomb's constant.

3. **Calculate Initial Potential Energy \( U_1 \)**:
- For \( r_1 = 0.2 \, m \):
\[ U_1 = k \frac{Q_1 Q_2}{r_1} = 9 \times 10^9 \frac{(5 \times 10^{-6})(4 \times 10^{-6})}{0.2} \]
- Simplifying:
\[ U_1 = 9 \times 10^9 \frac{20 \times 10^{-12}}{0.2} = 9 \times 10^9 \times 100 \times 10^{-12} = 9 \times 10^{-1} \, J = 0.9 \, J \]

4. **Calculate Final Potential Energy \( U_2 \)**:
- For \( r_2 = 0.1 \, m \):
\[ U_2 = k \frac{Q_1 Q_2}{r_2} = 9 \times 10^9 \frac{(5 \times 10^{-6})(4 \times 10^{-6})}{0.1} \]
- Simplifying:
\[ U_2 = 9 \times 10^9 \frac{20 \times 10^{-12}}{0.1} = 9 \times 10^9 \times 200 \times 10^{-12} = 18 \times 10^{-1} \, J = 1.8 \, J \]

5. **Calculate the Work Done \( W \)**:
- The work done to change the distance is given by:
\[ W = U_2 - U_1 \]
- Substituting the values:
\[ W = 1.8 \, J - 0.9 \, J = 0.9 \, J \]

### Final Answer:
The work done to decrease the distance between the two charges from 20 cm to 10 cm is **0.9 Joules**.