Charges `Q-q, Q,-q` are placed at the corners `A, B, C,D` of a square respectively. If the resultant force on the charge `Q` is zero due to other charges, what is the relation between `Q` and `q`?

Options

A. `Q= -2sqrt(2)q`
B. `Q= -2q`
C. `Q= -sqrt(2)q`
D. `Q= -(1)/(2sqrt(2))q`

Explanation

To find the relation between the charges \( Q \) and \( q \) such that the resultant force on the charge \( Q \) at corner \( A \) of the square is zero due to the other charges at corners \( B \) and \( C \), we can follow these steps:

### Step 1: Identify the charges and their positions
- Place the charges at the corners of a square:
- Charge \( Q \) at corner \( A \)
- Charge \( -q \) at corner \( B \)
- Charge \( Q \) at corner \( C \)
- Charge \( -q \) at corner \( D \)

### Step 2: Calculate the forces acting on charge \( Q \)
- The force on charge \( Q \) due to charge \( -q \) at \( B \) (denote this force as \( F_B \)):
\[ F_B = \frac{k \cdot |Q \cdot (-q)|}{A^2} = \frac{k \cdot Q \cdot q}{A^2} \]
This force acts towards charge \( -q \) at \( B \).

- The force on charge \( Q \) due to charge \( Q \) at \( C \) (denote this force as \( F_C \)):
\[ F_C = \frac{k \cdot |Q \cdot Q|}{A^2} = \frac{k \cdot Q^2}{A^2} \]
This force acts away from charge \( Q \) at \( C \).

### Step 3: Analyze the direction of the forces
- The force \( F_B \) acts towards \( B \) (downwards).
- The force \( F_C \) acts away from \( C \) (to the right).

### Step 4: Resolve the forces
- Since \( F_B \) and \( F_C \) are at right angles to each other, we can find the resultant force \( F_{BC} \) using the Pythagorean theorem:
\[ F_{BC} = \sqrt{F_B^2 + F_C^2} \]

### Step 5: Set the resultant force equal to the force due to charge \( Q \) at \( A \)
- For the resultant force on charge \( Q \) to be zero, the force \( F_A \) (which is the force due to the other charges) must equal the force \( F_{BC} \):
\[ F_A = F_{BC} \]

### Step 6: Substitute the expressions for forces
- Substitute the expressions for \( F_B \) and \( F_C \):
\[ \frac{k \cdot Q^2}{2A^2} = \sqrt{\left(\frac{k \cdot Q \cdot q}{A^2}\right)^2 + \left(\frac{k \cdot Q^2}{A^2}\right)^2} \]

### Step 7: Simplify the equation
- Square both sides to eliminate the square root:
\[ \left(\frac{k \cdot Q^2}{2A^2}\right)^2 = \left(\frac{k \cdot Q \cdot q}{A^2}\right)^2 + \left(\frac{k \cdot Q^2}{A^2}\right)^2 \]
- This leads to:
\[ \frac{k^2 \cdot Q^4}{4A^4} = \frac{k^2 \cdot Q^2 \cdot q^2}{A^4} + \frac{k^2 \cdot Q^4}{A^4} \]

### Step 8: Cancel common terms and solve for \( q \)
- Cancel \( k^2/A^4 \) from both sides:
\[ \frac{Q^4}{4} = Q^2 \cdot q^2 + Q^4 \]
- Rearranging gives:
\[ \frac{Q^4}{4} - Q^4 = Q^2 \cdot q^2 \]
- Simplifying:
\[ -\frac{3Q^4}{4} = Q^2 \cdot q^2 \]
- Dividing by \( Q^2 \) (assuming \( Q \neq 0 \)):
\[ q^2 = -\frac{3Q^2}{4} \]
- Taking the square root gives:
\[ q = \pm \sqrt{-\frac{3}{4}} Q = \pm \frac{\sqrt{3}}{2} Q \]

### Final Relation
Thus, the relation between \( Q \) and \( q \) is:
\[ q = \pm \frac{\sqrt{3}}{2} Q \]