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)/(4)`
B. `+Q`
C. `-Q`
D. `(Q)/(2)`

Answer

To find the value of charge \( q \) that will keep the system of three charges in equilibrium, we can follow these steps:

### Step 1: Understand the arrangement of charges
We have two equal charges \( Q \) placed at a distance \( L \) apart, and a charge \( q \) is placed at the midpoint of the line joining these two charges. The distances from \( q \) to each \( Q \) are \( \frac{L}{2} \).

### Step 2: Determine the forces acting on charge \( q \)
- The charge \( q \) will experience a repulsive force \( F_1 \) due to one of the charges \( Q \) and an attractive force \( F_2 \) due to the other charge \( Q \) if \( q \) is negative.

### Step 3: Calculate the forces
1. **Repulsive force \( F_1 \)** (from one \( Q \)):
\[ F_1 = k \frac{Q \cdot q}{\left(\frac{L}{2}\right)^2} = k \frac{Q \cdot q}{\frac{L^2}{4}} = \frac{4kQq}{L^2} \]

2. **Attractive force \( F_2 \)** (from the other \( Q \)):
\[ F_2 = k \frac{Q \cdot q}{\left(\frac{L}{2}\right)^2} = k \frac{Q \cdot q}{\frac{L^2}{4}} = \frac{4kQq}{L^2} \]

### Step 4: Set the forces equal for equilibrium
For the system to be in equilibrium, the magnitudes of the forces must be equal:
\[ F_1 = F_2 \]
Since both forces are equal in magnitude but opposite in direction, we can set them equal:
\[ \frac{4kQq}{L^2} = \frac{4kQ(-q)}{L^2} \]

### Step 5: Solve for \( q \)
Since both sides are equal, we can simplify:
\[ 4kQq = -4kQq \]
This implies:
\[ 4kQq + 4kQq = 0 \implies 8kQq = 0 \]
Since \( k \) and \( Q \) are not zero, we can divide both sides by \( 4kQ \):
\[ q = -\frac{Q}{4} \]

### Conclusion
Thus, the charge \( q \) must be equal to:
\[ q = -\frac{Q}{4} \]