Charges 4Q, q and Q and placed along x-axis at positions `x=0,x=t//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 of finding the value of charge \( q \) such that the net force on charge \( Q \) is zero, we can follow these steps:

### Step 1: Identify the positions of the charges
We have three charges:
- Charge \( 4Q \) is located at \( x = 0 \)
- Charge \( q \) is located at \( x = \frac{1}{2} \)
- Charge \( Q \) is located at \( x = 1 \)

### Step 2: Determine the forces acting on charge \( Q \)
The forces acting on charge \( Q \) are due to the other two charges:
1. The force \( F_1 \) due to charge \( 4Q \) (located at \( x = 0 \)).
2. The force \( F_2 \) due to charge \( q \) (located at \( x = \frac{1}{2} \)).

### Step 3: Calculate the force \( F_1 \) on charge \( Q \)
The force \( F_1 \) exerted by charge \( 4Q \) on charge \( Q \) is given by Coulomb's law:
\[ F_1 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{(4Q)(Q)}{(1-0)^2} = \frac{4Q^2}{4 \pi \epsilon_0} \]
This force is repulsive since both charges are positive.

### Step 4: Calculate the force \( F_2 \) on charge \( Q \)
The force \( F_2 \) exerted by charge \( q \) on charge \( Q \) is also given by Coulomb's law:
\[ F_2 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{(q)(Q)}{(1 - \frac{1}{2})^2} = \frac{qQ}{4 \pi \epsilon_0 \cdot \left(\frac{1}{2}\right)^2} = \frac{4qQ}{4 \pi \epsilon_0} \]
This force is attractive if \( q \) is negative (since \( Q \) is positive).

### Step 5: Set the net force on charge \( Q \) to zero
For the net force on charge \( Q \) to be zero, we need:
\[ F_1 - F_2 = 0 \]
Substituting the expressions for \( F_1 \) and \( F_2 \):
\[ \frac{4Q^2}{4 \pi \epsilon_0} - \frac{4qQ}{4 \pi \epsilon_0} = 0 \]
This simplifies to:
\[ 4Q^2 = 4qQ \]
Dividing both sides by \( 4Q \) (assuming \( Q \neq 0 \)):
\[ Q = q \]

### Step 6: Determine the sign of \( q \)
Since the force \( F_1 \) is repulsive and \( F_2 \) must be attractive to balance it, \( q \) must be negative:
\[ q = -Q \]

### Conclusion
Thus, the value of charge \( q \) that makes the net force on charge \( Q \) equal to zero is:
\[ q = -Q \]