Three charges `4q, Q` and `q` are in a straight line in the position of `0. l//2` and `l` respectively. The resultant force on `q` will be zero, if `Q=`

Options

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

Explanation

To solve the problem, we need to find the value of charge \( Q \) such that the resultant force on charge \( q \) is zero. We have three charges: \( 4q \) at position \( 0 \), \( Q \) at position \( \frac{l}{2} \), and \( q \) at position \( l \).

### Step-by-Step Solution:

1. **Identify the Positions of the Charges**:
- Charge \( 4q \) is at position \( 0 \).
- Charge \( Q \) is at position \( \frac{l}{2} \).
- Charge \( q \) is at position \( l \).

2. **Calculate the Forces Acting on Charge \( q \)**:
- The force on charge \( q \) due to charge \( 4q \) (denoted as \( F_{4q} \)):
\[ F_{4q} = k \cdot \frac{|4q \cdot q|}{(l)^2} = \frac{4kq^2}{l^2} \]
This force is directed towards the left (attracting \( q \) towards \( 4q \)).

- The force on charge \( q \) due to charge \( Q \) (denoted as \( F_Q \)):
\[ F_Q = k \cdot \frac{|Q \cdot q|}{\left(\frac{l}{2} - l\right)^2} = k \cdot \frac{|Q \cdot q|}{\left(-\frac{l}{2}\right)^2} = \frac{4kQq}{l^2} \]
This force is directed towards the right (repelling \( q \) away from \( Q \)).

3. **Set the Forces Equal for Equilibrium**:
For the resultant force on charge \( q \) to be zero, the magnitudes of the forces must be equal:
\[ F_{4q} = F_Q \]
Therefore,
\[ \frac{4kq^2}{l^2} = \frac{4kQq}{l^2} \]

4. **Simplify the Equation**:
We can cancel \( 4k \) and \( l^2 \) from both sides (assuming \( q \neq 0 \)):
\[ q = Q \]

5. **Conclusion**:
The value of charge \( Q \) must be:
\[ Q = -q \]
This means that for the resultant force on charge \( q \) to be zero, charge \( Q \) must be equal to \( -q \).

### Final Answer:
\[ Q = -q \]