A charges Q is placed at each of the two opposite corners of a square. A charge q is placed to each of the other two corners. If the resultant force on each charge q is zero, then

Options

A. `q=sqrt2Q`
B. `q=-sqrt2Q`
C. `q=2sqrt2Q`
D. `q=-2sqrt2Q`

Explanation

To solve the problem step by step, we will analyze the forces acting on the charges placed at the corners of the square.

### Step-by-Step Solution

1. **Understanding the Setup**:
- We have a square with side length \( a \).
- At two opposite corners of the square, we place charges \( Q \).
- At the other two opposite corners, we place charges \( q \).

2. **Identifying Forces on Charge \( q \)**:
- Each charge \( q \) experiences electrostatic forces due to the two charges \( Q \).
- The force \( F \) between each charge \( Q \) and charge \( q \) is given by Coulomb's law:
\[ F = k \frac{Qq}{a^2} \]
- Since there are two charges \( Q \) acting on each charge \( q \), we need to consider the vector nature of these forces.

3. **Calculating the Resultant Force**:
- The forces from the two charges \( Q \) on a charge \( q \) will act along the lines connecting the charges.
- The angle between the forces from the two \( Q \) charges acting on \( q \) is \( 90^\circ \) (since they are at right angles in the square).
- The resultant force \( F_{resultant} \) on charge \( q \) due to both \( Q \) charges can be calculated using the Pythagorean theorem:
\[ F_{resultant} = \sqrt{F^2 + F^2} = \sqrt{2F^2} = F\sqrt{2} \]

4. **Force from Charges \( q \)**:
- Each charge \( q \) also experiences a force due to the other charge \( q \) at the opposite corner.
- The distance between the two charges \( q \) is the diagonal of the square, which is \( \sqrt{2}a \).
- The force \( F' \) between the two charges \( q \) is given by:
\[ F' = k \frac{q^2}{(\sqrt{2}a)^2} = k \frac{q^2}{2a^2} \]

5. **Setting the Net Force to Zero**:
- For the resultant force on charge \( q \) to be zero, the force from the charges \( Q \) must balance the force from the other charge \( q \):
\[ F\sqrt{2} + F' = 0 \]
- Substituting the expressions for \( F \) and \( F' \):
\[ k \frac{Qq}{a^2} \sqrt{2} + k \frac{q^2}{2a^2} = 0 \]

6. **Simplifying the Equation**:
- Cancel \( k \) and \( a^2 \) from the equation:
\[ Qq\sqrt{2} + \frac{q^2}{2} = 0 \]
- Rearranging gives:
\[ Qq\sqrt{2} = -\frac{q^2}{2} \]

7. **Solving for \( q \)**:
- Dividing both sides by \( q \) (assuming \( q \neq 0 \)):
\[ Q\sqrt{2} = -\frac{q}{2} \]
- Therefore, we find:
\[ q = -2\sqrt{2}Q \]

### Final Answer:
The value of charge \( q \) is:
\[ q = -2\sqrt{2}Q \]