Charges `5 mu C, -2muC, 3muCand -9muC` are placed at the corners A,B,C and D of a square ABCD of side 1m. The net electric potential at the centre of th square is

Options

A. `-27 KV`
B. `-27 sqrt(2) KV`
C. `-90KV`
D. zero

Answer

To find the net electric potential at the center of the square formed by the charges at its corners, we will follow these steps:

### Step 1: Identify the charges and their positions
The charges are:
- At corner A: \( Q_A = 5 \, \mu C = 5 \times 10^{-6} \, C \)
- At corner B: \( Q_B = -2 \, \mu C = -2 \times 10^{-6} \, C \)
- At corner C: \( Q_C = 3 \, \mu C = 3 \times 10^{-6} \, C \)
- At corner D: \( Q_D = -9 \, \mu C = -9 \times 10^{-6} \, C \)

### Step 2: Calculate the distance from the center to each corner
For a square with a side length of \( 1 \, m \), the distance from the center of the square to any corner (using the Pythagorean theorem) is:
\[ d = \frac{\sqrt{1^2 + 1^2}}{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \, m \]

### Step 3: Write the formula for electric potential
The electric potential \( V \) due to a point charge \( Q \) at a distance \( r \) is given by:
\[ V = \frac{kQ}{r} \]
where \( k \) is Coulomb's constant, \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \).

### Step 4: Calculate the potential at the center due to each charge
The total potential at the center \( V_O \) is the sum of the potentials due to each charge:
\[ V_O = V_A + V_B + V_C + V_D \]
Substituting the values:
\[ V_O = \frac{k \cdot Q_A}{d} + \frac{k \cdot Q_B}{d} + \frac{k \cdot Q_C}{d} + \frac{k \cdot Q_D}{d} \]
Factoring out \( k/d \):
\[ V_O = \frac{k}{d} \left( Q_A + Q_B + Q_C + Q_D \right) \]

### Step 5: Substitute the values of the charges and distance
Substituting \( d = \frac{1}{\sqrt{2}} \):
\[ V_O = k \sqrt{2} \left( Q_A + Q_B + Q_C + Q_D \right) \]
Calculating the sum of the charges:
\[ Q_A + Q_B + Q_C + Q_D = 5 \times 10^{-6} - 2 \times 10^{-6} + 3 \times 10^{-6} - 9 \times 10^{-6} = (5 - 2 + 3 - 9) \times 10^{-6} = -3 \times 10^{-6} \, C \]

### Step 6: Substitute the values into the potential formula
Now substituting back:
\[ V_O = k \sqrt{2} \left( -3 \times 10^{-6} \right) \]
\[ V_O = 9 \times 10^9 \cdot \sqrt{2} \cdot (-3 \times 10^{-6}) \]
Calculating:
\[ V_O = -27 \sqrt{2} \times 10^3 \, V \]

### Final Answer
The net electric potential at the center of the square is:
\[ V_O = -27 \sqrt{2} \times 10^3 \, V \]