We may define electrostatic potential at a point in an electrostatic field as the amount of work done in moving a unit positive test charge from infinity to that point against the electrostatic forces, along any path. Due to a single charge `q` , potential at a point distant `r` from the charge is `V = (q)/(4pi in_(0)r)`. The potential can be positive or negative. However, it is scalar quantity. The total amount of work done in bringing various charges to their respective postions from infinelty large mutual separations gives us the electric potential energy of the system of charges. Whereas electric potentail is measured in volt, electric potential energy is measured in joule. You are given a square of each side 1.0 metre with four charges `+1xx10^(-8) C, -2xx10^(-8)C, +3xx10^(-8)C` and `+2xx10^(-8) C` placed at the four corners of the square. With the help of the passage given above, choose the most approprite alternative for each of the following questions :
Potential energy fo the system of four system of four charges is

Options

A. `12.73xx10^(7) J`
B. `-6.4xx10^(7) J`
C. `12.73xx10^(-9) J`
D. `-12.73xx10^(-9) J`

Explanation

To find the potential energy of the system of four charges placed at the corners of a square, we will follow these steps:

### Step 1: Identify the Charges and Their Positions
We have four charges located at the corners of a square with side length 1 meter:
- Charge \( q_1 = +1 \times 10^{-8} \, C \) at (0, 0)
- Charge \( q_2 = -2 \times 10^{-8} \, C \) at (1, 0)
- Charge \( q_3 = +3 \times 10^{-8} \, C \) at (1, 1)
- Charge \( q_4 = +2 \times 10^{-8} \, C \) at (0, 1)

### Step 2: Calculate the Potential Energy Between Each Pair of Charges
The potential energy \( U \) between two point charges \( q_i \) and \( q_j \) separated by a distance \( r \) is given by the formula:
\[ U_{ij} = k \frac{q_i q_j}{r} \]
where \( k = \frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^9 \, N \cdot m^2/C^2 \).

### Step 3: Calculate Pairwise Potential Energies
1. Between \( q_1 \) and \( q_2 \):
\[ U_{12} = k \frac{(1 \times 10^{-8})(-2 \times 10^{-8})}{1} = -2 \times 10^{-16} k \]

2. Between \( q_1 \) and \( q_3 \):
\[ U_{13} = k \frac{(1 \times 10^{-8})(3 \times 10^{-8})}{1} = 3 \times 10^{-16} k \]

3. Between \( q_1 \) and \( q_4 \):
\[ U_{14} = k \frac{(1 \times 10^{-8})(2 \times 10^{-8})}{1} = 2 \times 10^{-16} k \]

4. Between \( q_2 \) and \( q_3 \):
\[ U_{23} = k \frac{(-2 \times 10^{-8})(3 \times 10^{-8})}{1} = -6 \times 10^{-16} k \]

5. Between \( q_2 \) and \( q_4 \):
\[ U_{24} = k \frac{(-2 \times 10^{-8})(2 \times 10^{-8})}{1} = -8 \times 10^{-16} k \]

6. Between \( q_3 \) and \( q_4 \):
\[ U_{34} = k \frac{(3 \times 10^{-8})(2 \times 10^{-8})}{1} = 6 \times 10^{-16} k \]

### Step 4: Sum All Potential Energies
Now we sum all the potential energies calculated:
\[ U_{\text{total}} = U_{12} + U_{13} + U_{14} + U_{23} + U_{24} + U_{34} \]
Substituting the values:
\[ U_{\text{total}} = (-2 + 3 + 2 - 6 - 8 + 6) \times 10^{-16} k \]
\[ U_{\text{total}} = (-5) \times 10^{-16} k \]

### Step 5: Substitute \( k \) and Calculate
Substituting \( k = 9 \times 10^9 \):
\[ U_{\text{total}} = -5 \times 10^{-16} \times 9 \times 10^9 = -4.5 \times 10^{-6} \, J \]

### Step 6: Final Answer
The potential energy of the system of four charges is:
\[ U_{\text{total}} = -4.5 \times 10^{-6} \, J \]