Three charges each of `+4mu C` are the corners B, C, D of a square ABCD of side 1m. The electric field at the centre .O. of the square is

Options

A. `7.2 xx 10^(4) N//C` towards A
B. `7.2 xx 10^(4) N//C` towards C
C. `3.6 xx 10^(4) N//C` towards A
D. `3.6 xx 10^(4) N//C` towards C

Answer

To find the electric field at the center O of the square ABCD with three charges of +4 µC located at corners B, C, and D, we can follow these steps:

### Step 1: Understand the Configuration
We have a square ABCD with side length 1 m. The charges are placed at corners B, C, and D. The center O of the square is equidistant from all corners.

### Step 2: Determine the Distance from the Center to the Corners
The distance from the center O to any corner (like B, C, or D) can be calculated using the formula for the diagonal of a square. The diagonal (d) of a square with side length a is given by:
\[ d = a\sqrt{2} \]
Since we want the distance from the center to a corner, we take half of the diagonal:
\[ r = \frac{d}{2} = \frac{a\sqrt{2}}{2} \]
Substituting \( a = 1 \, \text{m} \):
\[ r = \frac{1 \cdot \sqrt{2}}{2} = \frac{\sqrt{2}}{2} \, \text{m} \]

### Step 3: Calculate the Electric Field Due to Each Charge
The electric field (E) due to a point charge is given by:
\[ E = \frac{k \cdot |q|}{r^2} \]
where \( k \) is Coulomb's constant (\( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)), \( q \) is the charge, and \( r \) is the distance from the charge to the point where we are calculating the field.

For each charge of \( +4 \, \mu C = 4 \times 10^{-6} \, C \):
\[ E = \frac{9 \times 10^9 \cdot 4 \times 10^{-6}}{(\frac{\sqrt{2}}{2})^2} \]
Calculating \( r^2 \):
\[ r^2 = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} = \frac{1}{2} \]
Thus, substituting \( r^2 \) into the electric field equation:
\[ E = \frac{9 \times 10^9 \cdot 4 \times 10^{-6}}{\frac{1}{2}} = 9 \times 10^9 \cdot 8 \times 10^{-6} = 72 \times 10^3 \, \text{N/C} \]

### Step 4: Determine the Direction of the Electric Fields
- The electric field due to charge at B points away from B towards the center O.
- The electric field due to charge at C points away from C towards the center O.
- The electric field due to charge at D points away from D towards the center O.

### Step 5: Analyze the Electric Fields at the Center
The electric fields due to charges at B and C will cancel each other out because they are equal in magnitude but opposite in direction. Therefore, we only need to consider the electric field due to charge D.

### Step 6: Resultant Electric Field at the Center
The resultant electric field at point O is simply the electric field due to charge D:
\[ E_{net} = E_D = 72 \times 10^3 \, \text{N/C} \]
This electric field will be directed towards point A (the opposite corner of D).

### Conclusion
The electric field at the center O of the square is:
\[ E = 72 \times 10^3 \, \text{N/C} \, \text{towards A} \]