10C and -10C are placed at y = 1 m and y=-1 m on y-axis 1c charge is placed on x-axis at x = +1. Now find the change in PE of system when 1 coulomb is displaced from x= +lm to x=-1 m keeping other two charges as fixed is

Options

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

Answer

To find the change in potential energy (PE) of the system when a 1 C charge is displaced from x = +1 m to x = -1 m while keeping the other two charges fixed, we can follow these steps:

### Step 1: Identify the Charges and Their Positions
- We have three charges:
- Charge \( q_1 = 10 \, C \) at \( (0, 1) \) m
- Charge \( q_2 = -10 \, C \) at \( (0, -1) \) m
- Charge \( q_3 = 1 \, C \) at \( (1, 0) \) m initially, which will be moved to \( (-1, 0) \) m.

### Step 2: Calculate Initial Potential Energy (U_initial)
The potential energy of the system can be calculated using the formula:
\[ U = k \sum \frac{q_i q_j}{r_{ij}} \]
where \( k \) is Coulomb's constant (\( k = 9 \times 10^9 \, N \cdot m^2/C^2 \)), \( q_i \) and \( q_j \) are the charges, and \( r_{ij} \) is the distance between the charges.

1. **Calculate \( U_{12} \) (between \( q_1 \) and \( q_2 \))**:
- Distance \( r_{12} = 2 \, m \)
\[ U_{12} = k \cdot \frac{q_1 \cdot q_2}{r_{12}} = 9 \times 10^9 \cdot \frac{10 \cdot (-10)}{2} = -45 \times 10^9 \, J \]

2. **Calculate \( U_{23} \) (between \( q_2 \) and \( q_3 \))**:
- Distance \( r_{23} = \sqrt{(0 - 1)^2 + (1 - (-1))^2} = \sqrt{1 + 4} = \sqrt{5} \, m \)
\[ U_{23} = k \cdot \frac{q_2 \cdot q_3}{r_{23}} = 9 \times 10^9 \cdot \frac{-10 \cdot 1}{\sqrt{5}} = -\frac{90 \times 10^9}{\sqrt{5}} \, J \]

3. **Calculate \( U_{13} \) (between \( q_1 \) and \( q_3 \))**:
- Distance \( r_{13} = \sqrt{(0 - 1)^2 + (1 - 0)^2} = \sqrt{1 + 1} = \sqrt{2} \, m \)
\[ U_{13} = k \cdot \frac{q_1 \cdot q_3}{r_{13}} = 9 \times 10^9 \cdot \frac{10 \cdot 1}{\sqrt{2}} = \frac{90 \times 10^9}{\sqrt{2}} \, J \]

4. **Total Initial Potential Energy**:
\[ U_{\text{initial}} = U_{12} + U_{23} + U_{13} \]

### Step 3: Calculate Final Potential Energy (U_final)
When the 1 C charge is moved to \( (-1, 0) \):
1. **Calculate \( U_{23} \) (between \( q_2 \) and \( q_3 \))**:
- Distance \( r_{23} = \sqrt{(0 - (-1))^2 + (1 - (-1))^2} = \sqrt{1 + 4} = \sqrt{5} \, m \)
\[ U_{23} = k \cdot \frac{-10 \cdot 1}{\sqrt{5}} = -\frac{90 \times 10^9}{\sqrt{5}} \, J \]

2. **Calculate \( U_{13} \) (between \( q_1 \) and \( q_3 \))**:
- Distance \( r_{13} = \sqrt{(0 - (-1))^2 + (1 - 0)^2} = \sqrt{1 + 1} = \sqrt{2} \, m \)
\[ U_{13} = k \cdot \frac{10 \cdot 1}{\sqrt{2}} = \frac{90 \times 10^9}{\sqrt{2}} \, J \]

3. **Total Final Potential Energy**:
\[ U_{\text{final}} = U_{12} + U_{23} + U_{13} \]

### Step 4: Calculate Change in Potential Energy
\[ \Delta U = U_{\text{final}} - U_{\text{initial}} \]
Since the configuration of the system remains the same and only the position of the 1 C charge changes, the potential energy contributions from the fixed charges do not change. Thus, we find that:
\[ \Delta U = 0 \]

### Conclusion
The change in potential energy of the system when the 1 C charge is displaced from \( x = +1 \, m \) to \( x = -1 \, m \) is:
\[ \Delta U = 0 \, J \]