Equal charges Q are placed at the four corners A, B, C, D of a square of length a. The magnitude of the force on the charge at B will be

Options

A. `(3Q^(2))/(4piepsilon_(0)a^(2))`
B. `(4Q^(2))/(4piepsilon_(0)a^(2))`
C. `((1+1sqrt(2))/(2))(Q^(2))/(4piepsilon_(0)a^(2))`
D. `(2+(1)/sqrt(2))(Q^(2))/(4piepsilon_(0)a^(2))`

Answer

To solve the problem of finding the magnitude of the force on the charge at corner B of a square with equal charges at each corner, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions**:
- Let the charges at corners A, B, C, and D be \( Q \).
- The square has a side length \( a \).

2. **Calculate the Forces Due to Each Charge on Charge B**:
- **Force due to Charge A (F_A)**:
- The distance between charges A and B is \( a \).
- The force \( F_A \) can be calculated using Coulomb's law:
\[ F_A = k \frac{Q^2}{a^2} \]
- This force acts along the line connecting A and B (to the right).

- **Force due to Charge C (F_C)**:
- The distance between charges B and C is also \( a \).
- The force \( F_C \) is given by:
\[ F_C = k \frac{Q^2}{a^2} \]
- This force acts along the line connecting B and C (upward).

- **Force due to Charge D (F_D)**:
- The distance between charges B and D is \( \sqrt{2}a \) (the diagonal of the square).
- The force \( F_D \) is given by:
\[ F_D = k \frac{Q^2}{( \sqrt{2}a )^2} = k \frac{Q^2}{2a^2} \]
- This force acts at a \( 45^\circ \) angle to the horizontal (downward and to the right).

3. **Resolve the Forces**:
- The forces \( F_A \) and \( F_C \) act along the x-axis and y-axis, respectively.
- The force \( F_D \) can be resolved into its x and y components:
- \( F_{Dx} = F_D \cos(45^\circ) = \frac{k Q^2}{2a^2} \cdot \frac{1}{\sqrt{2}} = \frac{k Q^2}{2\sqrt{2} a^2} \)
- \( F_{Dy} = F_D \sin(45^\circ) = \frac{k Q^2}{2a^2} \cdot \frac{1}{\sqrt{2}} = \frac{k Q^2}{2\sqrt{2} a^2} \)

4. **Calculate the Net Force in the X and Y Directions**:
- **Net Force in the X-Direction (F_x)**:
\[ F_x = F_A + F_{Dx} = k \frac{Q^2}{a^2} + \frac{k Q^2}{2\sqrt{2} a^2} \]
- **Net Force in the Y-Direction (F_y)**:
\[ F_y = F_C - F_{Dy} = k \frac{Q^2}{a^2} - \frac{k Q^2}{2\sqrt{2} a^2} \]

5. **Calculate the Magnitude of the Resultant Force**:
- The magnitude of the resultant force \( F_{net} \) can be found using the Pythagorean theorem:
\[ F_{net} = \sqrt{F_x^2 + F_y^2} \]

6. **Substituting the Values**:
- Substitute the expressions for \( F_x \) and \( F_y \) into the equation for \( F_{net} \) and simplify.

### Final Result:
The magnitude of the force on the charge at B can be expressed as:
\[ F_{net} = \sqrt{ \left( k \frac{Q^2}{a^2} + \frac{k Q^2}{2\sqrt{2} a^2} \right)^2 + \left( k \frac{Q^2}{a^2} - \frac{k Q^2}{2\sqrt{2} a^2} \right)^2 } \]