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 = sqrt2 q`
B. `Q = -sqrt2 q`
C. `Q = 2 sqrt2q`
D. `Q = - 2 sqrt2 q`

Answer

To solve the problem, we need to establish a relationship between the charges \( Q \) and \( q \) given that the resultant force on each charge \( q \) is zero. Let's break this down step by step.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have a square with charges \( Q \) at two opposite corners and charges \( q \) at the other two corners.
- Let's denote the corners of the square as \( A, B, C, D \) where \( A \) and \( C \) have charge \( Q \) and \( B \) and \( D \) have charge \( q \).

2. **Identifying Forces on Charge \( q \)**:
- The charge \( q \) at corner \( B \) experiences forces due to the two charges \( Q \) at corners \( A \) and \( C \), and the charge \( q \) at corner \( D \).
- The force between the two charges \( q \) (at corners \( B \) and \( D \)) is repulsive.

3. **Calculating the Force Between Charges**:
- The distance between the two charges \( q \) (at \( B \) and \( D \)) is the diagonal of the square, which is \( \sqrt{2}a \) (where \( a \) is the side length of the square).
- The force \( F_{qq} \) between the two charges \( q \) is given by Coulomb's law:
\[ F_{qq} = k \frac{q^2}{(\sqrt{2}a)^2} = k \frac{q^2}{2a^2} \]

4. **Calculating the Force Between Charge \( Q \) and Charge \( q \)**:
- The distance between charge \( Q \) (at \( A \) or \( C \)) and charge \( q \) (at \( B \)) is \( a \).
- The force \( F_{qQ} \) between charge \( Q \) and charge \( q \) is:
\[ F_{qQ} = k \frac{Qq}{a^2} \]

5. **Setting Up the Force Balance**:
- The net force on charge \( q \) at corner \( B \) must be zero. This means the attractive forces from \( Q \) must balance the repulsive force from the other \( q \):
\[ F_{qq} = 2F_{qQ} \]
- Since there are two \( Q \) charges exerting force on \( q \), we consider the resultant force from both \( Q \) charges. The resultant force from both \( Q \) charges acting on \( q \) at an angle of \( 45^\circ \) is:
\[ F_{resultant} = \sqrt{(F_{qQ})^2 + (F_{qQ})^2} = \sqrt{2} F_{qQ} \]

6. **Equating Forces**:
- Therefore, we have:
\[ k \frac{q^2}{2a^2} = \sqrt{2} \cdot k \frac{Qq}{a^2} \]
- Simplifying this gives:
\[ \frac{q^2}{2} = \sqrt{2} Qq \]

7. **Solving for \( q \)**:
- Rearranging the equation, we find:
\[ q^2 - 2\sqrt{2}Qq = 0 \]
- Factoring out \( q \):
\[ q(q - 2\sqrt{2}Q) = 0 \]
- This gives us two solutions: \( q = 0 \) or \( q = 2\sqrt{2}Q \).

### Final Relationship:
Thus, the relationship between the charges is:
\[ q = 2\sqrt{2}Q \]