10 C & -10C are placed at y = 1 & y = -1 m on y - axis. 1C charge is placed on x-axis at x = +1m. The change in PE of system when 1C is displaced from x = +1 to x = -1m keeping other two charges fixed is

Options

A. `10^(9)J`
B. `21xx10^(9)J`
C. `10xx10^(9)J`
D. zero

Explanation

To find the change in potential energy (PE) of the system when the 1 C charge is displaced from x = +1 m to x = -1 m, we will follow these steps:

### Step 1: Identify the initial configuration
- We have three charges:
- Charge \( Q_1 = +10 \, C \) at \( (0, 1) \)
- Charge \( Q_2 = -10 \, C \) at \( (0, -1) \)
- Charge \( Q_3 = +1 \, C \) initially at \( (1, 0) \)

### Step 2: Calculate the initial potential energy (PE_initial)
The potential energy of the system is given by the formula:
\[ PE = k \sum_{i < j} \frac{Q_i Q_j}{r_{ij}} \]
where \( k \) is Coulomb's constant (\( k \approx 8.99 \times 10^9 \, N \cdot m^2/C^2 \)), and \( r_{ij} \) is the distance between charges \( Q_i \) and \( Q_j \).

1. **Calculate the distance between charges:**
- Distance between \( Q_1 \) and \( Q_2 \):
\[ r_{12} = 2 \, m \]
- Distance between \( Q_1 \) and \( Q_3 \):
\[ r_{13} = \sqrt{(1-0)^2 + (0-1)^2} = \sqrt{2} \, m \]
- Distance between \( Q_2 \) and \( Q_3 \):
\[ r_{23} = \sqrt{(1-0)^2 + (0+1)^2} = \sqrt{2} \, m \]

2. **Calculate the potential energy contributions:**
- Between \( Q_1 \) and \( Q_2 \):
\[ PE_{12} = k \cdot \frac{10 \cdot (-10)}{2} = -50k \]
- Between \( Q_1 \) and \( Q_3 \):
\[ PE_{13} = k \cdot \frac{10 \cdot 1}{\sqrt{2}} = 5\sqrt{2}k \]
- Between \( Q_2 \) and \( Q_3 \):
\[ PE_{23} = k \cdot \frac{-10 \cdot 1}{\sqrt{2}} = -5\sqrt{2}k \]

3. **Total initial potential energy:**
\[ PE_{\text{initial}} = PE_{12} + PE_{13} + PE_{23} = -50k + 5\sqrt{2}k - 5\sqrt{2}k = -50k \]

### Step 3: Calculate the final potential energy (PE_final)
Now we will calculate the potential energy when the 1 C charge is moved to \( x = -1 \).

1. **Calculate the distances for the new position:**
- Distance between \( Q_1 \) and \( Q_3 \):
\[ r'_{13} = \sqrt{(-1-0)^2 + (0-1)^2} = \sqrt{2} \, m \]
- Distance between \( Q_2 \) and \( Q_3 \):
\[ r'_{23} = \sqrt{(-1-0)^2 + (0+1)^2} = \sqrt{2} \, m \]

2. **Calculate the potential energy contributions for the new position:**
- Between \( Q_1 \) and \( Q_2 \) (remains the same):
\[ PE_{12} = -50k \]
- Between \( Q_1 \) and \( Q_3 \):
\[ PE'_{13} = k \cdot \frac{10 \cdot 1}{\sqrt{2}} = 5\sqrt{2}k \]
- Between \( Q_2 \) and \( Q_3 \):
\[ PE'_{23} = k \cdot \frac{-10 \cdot 1}{\sqrt{2}} = -5\sqrt{2}k \]

3. **Total final potential energy:**
\[ PE_{\text{final}} = PE_{12} + PE'_{13} + PE'_{23} = -50k + 5\sqrt{2}k - 5\sqrt{2}k = -50k \]

### Step 4: Calculate the change in potential energy
\[ \Delta PE = PE_{\text{final}} - PE_{\text{initial}} = -50k - (-50k) = 0 \]

### Final Answer
The change in potential energy of the system when the 1 C charge is displaced from \( x = +1 \) m to \( x = -1 \) m is **0**.

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