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

Answer

To find the magnitude of the force on the charge at point B due to the equal charges placed at the corners of a square, we will follow these steps:

### Step 1: Understand the Configuration
We have a square ABCD with equal charges \( Q \) placed at each corner. We need to determine the net force acting on the charge at corner B due to the charges at corners A, C, and D.

### Step 2: Calculate the Force from Charge A
The distance between charges A and B is equal to the side length of the square, which is \( a \). The force \( F_{AB} \) exerted on charge B by charge A is given by Coulomb's law:
\[ F_{AB} = \frac{k Q^2}{a^2} \]
where \( k \) is Coulomb's constant.

### Step 3: Calculate the Force from Charge D
Similarly, the distance between charges D and B is also \( a \). The force \( F_{DB} \) exerted on charge B by charge D is:
\[ F_{DB} = \frac{k Q^2}{a^2} \]

### Step 4: Calculate the Force from Charge C
The distance between charges C and B is the diagonal of the square. Using the Pythagorean theorem, the distance \( r_{BC} \) is:
\[ r_{BC} = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2} \]
The force \( F_{CB} \) exerted on charge B by charge C is:
\[ F_{CB} = \frac{k Q^2}{(a\sqrt{2})^2} = \frac{k Q^2}{2a^2} \]

### Step 5: Resolve Forces into Components
Next, we need to resolve the forces \( F_{AB} \) and \( F_{DB} \) into their x and y components. Since both forces \( F_{AB} \) and \( F_{DB} \) act along the axes:
- \( F_{AB} \) acts in the negative y-direction.
- \( F_{DB} \) acts in the positive y-direction.

The force \( F_{CB} \) has both x and y components:
- The x-component of \( F_{CB} \) is:
\[ F_{CBx} = F_{CB} \cos(45^\circ) = \frac{k Q^2}{2a^2} \cdot \frac{1}{\sqrt{2}} = \frac{k Q^2}{2\sqrt{2} a^2} \]
- The y-component of \( F_{CB} \) is:
\[ F_{CBy} = F_{CB} \sin(45^\circ) = \frac{k Q^2}{2a^2} \cdot \frac{1}{\sqrt{2}} = \frac{k Q^2}{2\sqrt{2} a^2} \]

### Step 6: Calculate Net Force in x and y Directions
Now we can calculate the net forces in the x and y directions:
- **Net force in the x-direction**:
\[ F_{net,x} = F_{CBx} = \frac{k Q^2}{2\sqrt{2} a^2} \]
- **Net force in the y-direction**:
\[ F_{net,y} = F_{DB} - F_{AB} + F_{CBy} = \frac{k Q^2}{a^2} - \frac{k Q^2}{a^2} + \frac{k Q^2}{2\sqrt{2} a^2} = \frac{k Q^2}{2\sqrt{2} a^2} \]

### Step 7: Calculate the Magnitude of the Resultant Force
To find the magnitude of the resultant force \( F_{net} \):
\[ F_{net} = \sqrt{(F_{net,x})^2 + (F_{net,y})^2} \]
Substituting the values:
\[ F_{net} = \sqrt{\left(\frac{k Q^2}{2\sqrt{2} a^2}\right)^2 + \left(\frac{k Q^2}{2\sqrt{2} a^2}\right)^2} \]
\[ F_{net} = \sqrt{2 \left(\frac{k Q^2}{2\sqrt{2} a^2}\right)^2} = \frac{k Q^2}{2\sqrt{2} a^2} \sqrt{2} = \frac{k Q^2}{2 a^2} \]

### Final Result
Thus, the magnitude of the force on the charge at B is:
\[ F_{net} = \frac{k Q^2}{2 a^2} \]