A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to:

Options

A. `Q//2`
B. `-Q//2`
C. `Q//4`
D. `-Q//4`

Answer

To solve the problem, we need to find the value of the charge \( q \) that will ensure the equilibrium of the system consisting of two equal charges \( Q \) and a charge \( q \) placed at the center of the line joining them.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two equal charges \( Q \) placed at a distance \( 2r \) apart.
- The charge \( q \) is placed at the midpoint, which is at a distance \( r \) from each charge \( Q \).

2. **Forces Acting on Charge \( q \)**:
- The charge \( q \) will experience electrostatic forces due to both charges \( Q \).
- The force exerted on \( q \) by one charge \( Q \) is attractive if \( Q \) is positive and \( q \) is negative (and vice versa).

3. **Calculating Forces**:
- The force \( F_Q \) on charge \( q \) due to one charge \( Q \) is given by Coulomb's law:
\[ F_Q = k \frac{Q \cdot |q|}{r^2} \]
- Since there are two charges \( Q \), the total force on \( q \) will be the sum of the forces due to both charges \( Q \).

4. **Equilibrium Condition**:
- For the system to be in equilibrium, the net force acting on charge \( q \) must be zero.
- This means that the force due to the left charge \( Q \) must equal the force due to the right charge \( Q \):
\[ F_{Q1} + F_{Q2} = 0 \]

5. **Setting Up the Equation**:
- The force on \( q \) due to the left charge \( Q \) (let's call it \( F_{Q1} \)) is attractive:
\[ F_{Q1} = k \frac{Q \cdot |q|}{r^2} \]
- The force on \( q \) due to the right charge \( Q \) (let's call it \( F_{Q2} \)) is also attractive:
\[ F_{Q2} = k \frac{Q \cdot |q|}{r^2} \]
- Since both forces are in the same direction (towards the charges \( Q \)), we can write:
\[ F_{Q1} = -F_{Q2} \]

6. **Finding the Value of \( q \)**:
- The force \( F_{Q} \) acting on charge \( Q \) due to charge \( q \) must balance the forces acting on \( Q \):
\[ k \frac{Q \cdot |q|}{r^2} = k \frac{q^2}{(2r)^2} \]
- Simplifying this gives:
\[ Q \cdot |q| = \frac{q^2}{4} \]
- Rearranging gives:
\[ 4Q \cdot |q| = q^2 \]
- This can be rearranged to:
\[ q^2 - 4Q \cdot |q| = 0 \]
- Factoring out \( q \):
\[ q(q - 4Q) = 0 \]
- This gives us two solutions: \( q = 0 \) or \( q = 4Q \).

7. **Considering the Sign of \( q \)**:
- Since \( q \) must be opposite in sign to \( Q \) for attraction, we take:
\[ q = -\frac{Q}{4} \]

8. **Final Answer**:
- Therefore, the charge \( q \) that will ensure equilibrium is:
\[ q = -\frac{Q}{4} \]

### Conclusion:
The system of three charges will be in equilibrium if \( q \) is equal to \( -\frac{Q}{4} \).