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

Answer

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

### Step 1: Understand the Setup
We have four charges, each of value \( Q = \frac{10}{3} \times 10^{-9} \, \text{C} \), placed at the corners of a square with a side length of \( 8 \, \text{cm} \).

### Step 2: Convert Units
Convert the side length from centimeters to meters:
\[ \text{Side length} = 8 \, \text{cm} = 8 \times 10^{-2} \, \text{m} \]

### Step 3: Calculate the Diagonal
The diagonal \( D \) of the square can be calculated using the Pythagorean theorem:
\[ D = \sqrt{(\text{side})^2 + (\text{side})^2} = \sqrt{(8 \times 10^{-2})^2 + (8 \times 10^{-2})^2} \]
\[ D = \sqrt{2 \times (8 \times 10^{-2})^2} = \sqrt{2 \times 64 \times 10^{-4}} = \sqrt{128 \times 10^{-4}} = 8\sqrt{2} \times 10^{-2} \, \text{m} \]

### Step 4: Find the Distance from the Center to a Corner
The distance \( R \) from the center of the square (intersection of the diagonals) to any corner is half of the diagonal:
\[ R = \frac{D}{2} = \frac{8\sqrt{2} \times 10^{-2}}{2} = 4\sqrt{2} \times 10^{-2} \, \text{m} \]

### Step 5: Calculate the Potential Due to One Charge
The electric potential \( V \) due to a single charge \( Q \) at a distance \( R \) is given by:
\[ V = \frac{kQ}{R} \]
where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

Substituting the values:
\[ V = \frac{9 \times 10^9 \times \left(\frac{10}{3} \times 10^{-9}\right)}{4\sqrt{2} \times 10^{-2}} \]

### Step 6: Calculate the Total Potential
Since there are four identical charges, the total potential \( V_{\text{total}} \) at the intersection point is:
\[ V_{\text{total}} = 4 \times V \]
\[ V_{\text{total}} = 4 \times \frac{9 \times 10^9 \times \left(\frac{10}{3} \times 10^{-9}\right)}{4\sqrt{2} \times 10^{-2}} \]
\[ V_{\text{total}} = \frac{9 \times 10^9 \times \left(\frac{10}{3} \times 10^{-9}\right)}{\sqrt{2} \times 10^{-2}} \]

### Step 7: Simplify the Expression
Calculating the above expression:
\[ V_{\text{total}} = \frac{9 \times 10^9 \times \frac{10}{3} \times 10^{-9}}{\sqrt{2} \times 10^{-2}} = \frac{90 \times 10^0}{3\sqrt{2}} = \frac{30}{\sqrt{2}} \, \text{V} \]
\[ V_{\text{total}} = 30\sqrt{2} \, \text{V} \]

### Final Answer
Thus, the potential at the intersection of the diagonals is:
\[ V_{\text{total}} = 1500\sqrt{2} \, \text{V} \]