A charge Q is place at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then `Q//q` equals:

Options

A. `-2 sqrt2`
B. `-1`
C. 1
D. `-(1)/(sqrt2)`

Answer

To solve the problem, we need to analyze the forces acting on the charge \( Q \) placed at one corner of the square due to the other charges.

### Step-by-step Solution:

1. **Identify the Charges and Their Positions**:
- Place charge \( Q \) at corners A and C of the square.
- Place charge \( q \) at corners B and D.

2. **Determine the Forces Acting on Charge \( Q \)**:
- The force on charge \( Q \) at corner A due to the charge \( Q \) at corner C (diagonal) is given by:
\[ F_1 = k \frac{Q^2}{(a\sqrt{2})^2} = k \frac{Q^2}{2a^2} \]
- The forces on charge \( Q \) at corner A due to the charges \( q \) at corners B and D (adjacent) are:
\[ F_2 = k \frac{Qq}{a^2} \quad \text{(from charge at B)} \]
\[ F_3 = k \frac{Qq}{a^2} \quad \text{(from charge at D)} \]

3. **Calculate the Resultant Force from Charges \( q \)**:
- The forces \( F_2 \) and \( F_3 \) act at 90 degrees to each other. The resultant force \( F_R \) can be calculated using the Pythagorean theorem:
\[ F_R = \sqrt{F_2^2 + F_3^2} = \sqrt{\left(k \frac{Qq}{a^2}\right)^2 + \left(k \frac{Qq}{a^2}\right)^2} = \sqrt{2} \cdot k \frac{Qq}{a^2} \]

4. **Set the Net Force on Charge \( Q \) to Zero**:
- For the net force on charge \( Q \) to be zero, the force due to the other charge \( Q \) must balance the resultant force from the charges \( q \):
\[ F_1 - F_R = 0 \]
\[ k \frac{Q^2}{2a^2} = \sqrt{2} \cdot k \frac{Qq}{a^2} \]

5. **Simplify the Equation**:
- Cancel \( k \) and \( a^2 \) from both sides:
\[ \frac{Q^2}{2} = \sqrt{2} Qq \]
- Rearranging gives:
\[ Q^2 = 2\sqrt{2} Qq \]

6. **Solve for the Ratio \( \frac{Q}{q} \)**:
- Dividing both sides by \( Q \) (assuming \( Q \neq 0 \)):
\[ Q = 2\sqrt{2} q \]
- Therefore, the ratio \( \frac{Q}{q} \) is:
\[ \frac{Q}{q} = 2\sqrt{2} \]

### Final Answer:
\[ \frac{Q}{q} = 2\sqrt{2} \]