Two point charges of `+ 1.0 mu C ` are kept stationary 2m apart. How much work is needed to be done to bring them 1m apart ?

Options

A. 4.5 mj
B. 9 mj
C. 45 mj
D. 90 mj

Answer

To solve the problem of how much work is needed to bring two point charges of +1.0 µC from a distance of 2 meters apart to 1 meter apart, we can follow these steps:

### Step 1: Understand the Concept of Work Done
The work done (W) in bringing the charges closer is equal to the change in potential energy (ΔPE) of the system. This can be expressed as:
\[ W = \Delta PE = PE_{final} - PE_{initial} \]

### Step 2: Calculate Initial Potential Energy
The potential energy (PE) between two point charges is given by the formula:
\[ PE = \frac{k \cdot q_1 \cdot q_2}{r} \]
where:
- \( k \) is Coulomb's constant (\( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \))
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges
- \( r \) is the distance between the charges

For the initial configuration (2 meters apart):
\[ PE_{initial} = \frac{k \cdot q_1 \cdot q_2}{r_i} = \frac{9 \times 10^9 \cdot (1 \times 10^{-6}) \cdot (1 \times 10^{-6})}{2} \]

### Step 3: Substitute Values for Initial Potential Energy
Substituting the values:
\[ PE_{initial} = \frac{9 \times 10^9 \cdot 1 \times 10^{-12}}{2} = \frac{9 \times 10^{-3}}{2} = 4.5 \times 10^{-3} \, \text{J} \]

### Step 4: Calculate Final Potential Energy
Now, calculate the potential energy when the charges are 1 meter apart:
\[ PE_{final} = \frac{k \cdot q_1 \cdot q_2}{r_f} = \frac{9 \times 10^9 \cdot (1 \times 10^{-6}) \cdot (1 \times 10^{-6})}{1} \]

### Step 5: Substitute Values for Final Potential Energy
Substituting the values:
\[ PE_{final} = 9 \times 10^9 \cdot 1 \times 10^{-12} = 9 \times 10^{-3} \, \text{J} \]

### Step 6: Calculate the Change in Potential Energy
Now, we can find the change in potential energy:
\[ \Delta PE = PE_{final} - PE_{initial} = 9 \times 10^{-3} - 4.5 \times 10^{-3} = 4.5 \times 10^{-3} \, \text{J} \]

### Step 7: Conclusion
The work done to bring the two charges from 2 meters apart to 1 meter apart is:
\[ W = 4.5 \times 10^{-3} \, \text{J} = 4.5 \, \text{mJ} \]

Thus, the answer is **4.5 mJ**.

---