Three charges, each equal to q, are placed at the three. corners of a square of side a . Find the electric field at. the fourth corner.

Explanation

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

### Step 1: Identify the Charges and Their Positions
We have three charges, each equal to \( q \), placed at the corners A, B, and C of a square with side length \( a \). The fourth corner, D, is where we want to find the electric field.

### Step 2: Calculate the Electric Field Due to Each Charge
The electric field \( E \) due to a point charge \( q \) at a distance \( r \) is given by the formula:
\[ E = \frac{k \cdot q}{r^2} \]
where \( k \) is Coulomb's constant.

- **Electric Field at D due to Charge at A (E1)**:
The distance from A to D is \( a \).
\[ E_1 = \frac{k \cdot q}{a^2} \]

- **Electric Field at D due to Charge at B (E2)**:
The distance from B to D is \( a \).
\[ E_2 = \frac{k \cdot q}{a^2} \]

- **Electric Field at D due to Charge at C (E3)**:
The distance from C to D is the diagonal of the square, which is \( a\sqrt{2} \).
\[ E_3 = \frac{k \cdot q}{(a\sqrt{2})^2} = \frac{k \cdot q}{2a^2} \]

### Step 3: Determine the Directions of the Electric Fields
- **E1** points directly away from A towards D.
- **E2** points directly away from B towards D.
- **E3** points directly away from C towards D.

### Step 4: Resolve the Electric Fields into Components
- **E1** (along the x-axis):
\[ E_{1x} = E_1 = \frac{k \cdot q}{a^2}, \quad E_{1y} = 0 \]

- **E2** (along the y-axis):
\[ E_{2x} = 0, \quad E_{2y} = E_2 = \frac{k \cdot q}{a^2} \]

- **E3** (diagonal):
The components of \( E_3 \) can be calculated as:
\[ E_{3x} = E_3 \cdot \cos(45^\circ) = \frac{k \cdot q}{2a^2} \cdot \frac{1}{\sqrt{2}} = \frac{k \cdot q}{2a^2\sqrt{2}} \]
\[ E_{3y} = E_3 \cdot \sin(45^\circ) = \frac{k \cdot q}{2a^2} \cdot \frac{1}{\sqrt{2}} = \frac{k \cdot q}{2a^2\sqrt{2}} \]

### Step 5: Sum the Components of the Electric Fields
Now, we can sum the components in the x and y directions:

- **Net Electric Field in the x-direction**:
\[ E_{net,x} = E_{1x} + E_{3x} = \frac{k \cdot q}{a^2} + \frac{k \cdot q}{2a^2\sqrt{2}} \]

- **Net Electric Field in the y-direction**:
\[ E_{net,y} = E_{2y} + E_{3y} = \frac{k \cdot q}{a^2} + \frac{k \cdot q}{2a^2\sqrt{2}} \]

### Step 6: Calculate the Magnitude of the Resultant Electric Field
The magnitude of the resultant electric field \( E_{net} \) can be calculated using the Pythagorean theorem:
\[ E_{net} = \sqrt{E_{net,x}^2 + E_{net,y}^2} \]

### Final Expression for the Electric Field
Combining the results, we can express the net electric field at point D as:
\[ E_{net} = k \cdot q \left( \frac{2\sqrt{2} + 1}{2a^2} \right) \]