Point charge of `3xx10^(-9)C` are situated at each of three corners of a square whose side is `15cm`. The magnitude and direction of electric field at the vacant corner of the square is

Options

A. `2296V//m` along the diagonal
B. `9622V//m` along the diagonal
C. `22.0V//m` along the diagonal
D. Zero

Explanation

To solve the problem of finding the electric field at the vacant corner of a square with three point charges at the other corners, we can follow these steps:

### Step 1: Understand the Configuration
We have a square with side length \( a = 15 \, \text{cm} = 0.15 \, \text{m} \). There are three point charges of \( q = 3 \times 10^{-9} \, \text{C} \) located at three corners of the square. Let's label the corners as A, B, C, and D, where D is the vacant corner.

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

#### Electric Field at Corner D due to Charge at A (E_A)
- Distance \( r_{AD} = a = 0.15 \, \text{m} \)
\[ E_A = \frac{9 \times 10^9 \times 3 \times 10^{-9}}{(0.15)^2} = \frac{27 \times 10^0}{0.0225} = 1200 \, \text{N/C} \]
- Direction: Away from charge A (towards corner D).

#### Electric Field at Corner D due to Charge at B (E_B)
- Distance \( r_{BD} = a = 0.15 \, \text{m} \)
\[ E_B = \frac{9 \times 10^9 \times 3 \times 10^{-9}}{(0.15)^2} = 1200 \, \text{N/C} \]
- Direction: Away from charge B (towards corner D).

#### Electric Field at Corner D due to Charge at C (E_C)
- Distance \( r_{CD} = \sqrt{2}a = 0.15\sqrt{2} \, \text{m} \)
\[ E_C = \frac{9 \times 10^9 \times 3 \times 10^{-9}}{(0.15\sqrt{2})^2} = \frac{27 \times 10^0}{0.045} = 600 \, \text{N/C} \]
- Direction: Away from charge C (towards corner D).

### Step 3: Resolve the Electric Fields into Components
Since the electric fields due to charges at A and B are directed along the sides of the square, and the electric field due to charge C is directed diagonally, we can resolve these vectors.

- \( E_A \) and \( E_B \) are along the x and y axes respectively:
- \( E_{Ax} = 1200 \, \text{N/C} \) (to the right)
- \( E_{By} = 1200 \, \text{N/C} \) (upward)

- The electric field \( E_C \) can be resolved into components:
\[ E_{Cx} = E_C \cdot \frac{1}{\sqrt{2}} = 600 \cdot \frac{1}{\sqrt{2}} = 300\sqrt{2} \, \text{N/C} \]
\[ E_{Cy} = E_C \cdot \frac{1}{\sqrt{2}} = 600 \cdot \frac{1}{\sqrt{2}} = 300\sqrt{2} \, \text{N/C} \]

### Step 4: Sum the Components
Now we can sum the x and y components:
- Total \( E_x = E_{Ax} + E_{Cx} = 1200 + 300\sqrt{2} \)
- Total \( E_y = E_{By} + E_{Cy} = 1200 + 300\sqrt{2} \)

### Step 5: Calculate the Resultant Electric Field
The magnitude of the resultant electric field \( E \) can be calculated using the Pythagorean theorem:
\[ E = \sqrt{(E_x)^2 + (E_y)^2} \]

### Step 6: Calculate the Magnitude and Direction
Substituting the values:
\[ E_x = 1200 + 300\sqrt{2} \approx 1200 + 424.26 \approx 1624.26 \, \text{N/C} \]
\[ E_y = 1200 + 300\sqrt{2} \approx 1624.26 \, \text{N/C} \]
\[ E = \sqrt{(1624.26)^2 + (1624.26)^2} = 1624.26\sqrt{2} \approx 2297.0 \, \text{N/C} \]

### Step 7: Determine the Direction
The direction of the resultant electric field is along the diagonal of the square, pointing from the center towards the vacant corner D.

### Final Answer
The magnitude of the electric field at corner D is approximately \( 2297 \, \text{N/C} \) directed along the diagonal towards corner D.

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