Charges 4Q, q and Q and placed along x-axis at positions `x=0,x=1//2` and `x=1`, respectively. Find the value of q so that force on charge Q is zero

Options

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

Answer

To solve the problem, we need to find the value of charge \( q \) such that the net force acting on charge \( Q \) is zero. The charges are placed along the x-axis at the following positions:

- Charge \( 4Q \) is at \( x = 0 \)
- Charge \( q \) is at \( x = \frac{1}{2} \)
- Charge \( Q \) is at \( x = 1 \)

### Step-by-Step Solution:

1. **Identify the Forces Acting on Charge \( Q \)**:
- The force \( F_1 \) due to charge \( 4Q \) on charge \( Q \) is repulsive (since both charges are positive).
- The force \( F_2 \) due to charge \( q \) on charge \( Q \) will be attractive if \( q \) is negative (since \( Q \) is positive).

2. **Calculate the Magnitude of the Forces**:
- The distance between \( 4Q \) and \( Q \) is \( 1 \) unit. Thus, the force \( F_1 \) is given by:
\[ F_1 = k \frac{(4Q)(Q)}{(1)^2} = 4kQ^2 \]
- The distance between \( q \) and \( Q \) is \( \frac{1}{2} \) unit. Thus, the force \( F_2 \) is given by:
\[ F_2 = k \frac{(q)(Q)}{\left(\frac{1}{2}\right)^2} = k \frac{(q)(Q)}{\frac{1}{4}} = 4kqQ \]

3. **Set the Forces Equal to Each Other**:
- For the net force on charge \( Q \) to be zero, we set the magnitudes of the forces equal:
\[ F_1 = F_2 \]
\[ 4kQ^2 = 4kqQ \]

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

5. **Solve for \( q \)**:
- Dividing both sides by \( Q \) (assuming \( Q \neq 0 \)):
\[ q = Q \]

6. **Determine the Sign of \( q \)**:
- Since the force \( F_1 \) is repulsive, \( F_2 \) must be attractive to balance it. This means \( q \) must be negative. Therefore, we conclude:
\[ q = -Q \]

### Final Answer:
The value of \( q \) such that the force on charge \( Q \) is zero is:
\[ q = -Q \]