Charges of `+(10)/(3) xx 10^(-9) C` are placed at each of the four corners of a square of side `8 cm`. The potential at the intersection of the diagonals is

Options

A. `150 sqrt(2)` volt
B. `1500 sqrt(2)` volt
C. `900 sqrt(2)` volt
D. 900 volt

Explanation

To find the electric potential at the intersection of the diagonals of a square with charges placed at each corner, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Charge at each corner, \( Q = \frac{10}{3} \times 10^{-9} \, \text{C} \)
- Side of the square, \( a = 8 \, \text{cm} = 0.08 \, \text{m} \)

2. **Determine the Distance from the Corner to the Center:**
- The intersection of the diagonals is the center of the square. The distance from a corner to the center (diagonal distance) can be calculated using the formula:
\[ L = \frac{a}{\sqrt{2}} \]
- Substituting the value of \( a \):
\[ L = \frac{0.08}{\sqrt{2}} = \frac{0.08}{1.414} \approx 0.05657 \, \text{m} \]

3. **Calculate the Electric Potential at the Center:**
- The electric potential \( V \) due to a point charge is given by:
\[ V = \frac{kQ}{r} \]
- Where \( k \) is Coulomb's constant, \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).
- Since there are four charges, the total potential at the center is:
\[ V_0 = 4 \times \frac{kQ}{L} \]

4. **Substituting the Values:**
- Now substituting \( k \), \( Q \), and \( L \):
\[ V_0 = 4 \times \frac{9 \times 10^9 \times \left(\frac{10}{3} \times 10^{-9}\right)}{0.05657} \]
- Simplifying:
\[ V_0 = 4 \times \frac{9 \times \frac{10}{3} \times 10^9 \times 10^{-9}}{0.05657} \]
\[ V_0 = 4 \times \frac{30 \times 10^0}{0.05657} \]
\[ V_0 = 4 \times \frac{30}{0.05657} \approx 4 \times 530.5 \approx 2122 \, \text{V} \]

5. **Final Calculation:**
- The final potential at the intersection of the diagonals is approximately:
\[ V_0 \approx 2122 \, \text{V} \]

### Conclusion:
The potential at the intersection of the diagonals of the square is approximately **2122 V**.