Two point electric charges of value `q` and `2q` are kept at a distance `d` apart from each other in air. A third charge `Q` is to be kept along the same line in such a way that the net force action on `q` and `2q` is zero. Calculate the position of charge `Q` in terms of `q` and `d`.

Options

A. `(d)/(sqrt(2)-1)`
B. `(d)/(sqrt(2)+1)`
C. `(d)/(sqrt(3)+1)`
D. `(d)/(sqrt(3)-1)`

Explanation

To solve the problem of finding the position of charge \( Q \) such that the net force on charges \( q \) and \( 2q \) is zero, we can follow these steps:

### Step 1: Understand the Setup
We have two charges:
- Charge \( q \) is at position \( A \).
- Charge \( 2q \) is at position \( B \), which is \( d \) units away from \( A \).

We need to place a third charge \( Q \) along the line connecting \( A \) and \( B \) such that the net force on both \( q \) and \( 2q \) is zero.

### Step 2: Identify Possible Positions for Charge \( Q \)
Charge \( Q \) can be placed in three possible regions:
1. To the left of charge \( q \).
2. Between charges \( q \) and \( 2q \).
3. To the right of charge \( 2q \).

### Step 3: Analyze Each Position
1. **Left of \( q \)**:
- If \( Q \) is negative, it will attract \( q \) and repel \( 2q \). The net force cannot be zero.

2. **Between \( q \) and \( 2q \)**:
- If \( Q \) is negative, it will attract \( q \) and repel \( 2q \). The forces will not balance out.

3. **Right of \( 2q \)**:
- If \( Q \) is negative, it will attract \( 2q \) and repel \( q \). This could potentially balance the forces.

### Step 4: Set Up the Equations
Assume \( Q \) is placed at a distance \( x \) from charge \( q \) (thus \( d - x \) from charge \( 2q \)).

#### For Charge \( q \):
The force on \( q \) due to \( Q \) (attractive) and \( 2q \) (repulsive) can be expressed as:
\[ F_{q} = k \frac{|Q| \cdot |q|}{x^2} = k \frac{2q^2}{d^2} \]

#### For Charge \( 2q \):
The force on \( 2q \) due to \( Q \) (attractive) and \( q \) (repulsive) can be expressed as:
\[ F_{2q} = k \frac{|Q| \cdot |2q|}{(d-x)^2} = k \frac{q \cdot 2q}{d^2} \]

### Step 5: Equate the Forces
Set the magnitudes of the forces equal to each other:
\[ k \frac{|Q| \cdot |q|}{x^2} = k \frac{2q^2}{d^2} \]
\[ k \frac{|Q| \cdot |2q|}{(d-x)^2} = k \frac{q \cdot 2q}{d^2} \]

### Step 6: Solve the Equations
From the first equation:
\[ \frac{|Q|}{x^2} = \frac{2q}{d^2} \Rightarrow |Q| = \frac{2q x^2}{d^2} \]

From the second equation:
\[ \frac{|Q|}{(d-x)^2} = \frac{q}{d^2} \Rightarrow |Q| = \frac{q (d-x)^2}{d^2} \]

### Step 7: Set the Two Expressions for \( |Q| \) Equal
\[ \frac{2q x^2}{d^2} = \frac{q (d-x)^2}{d^2} \]
Cancelling \( q \) (assuming \( q \neq 0 \)):
\[ 2x^2 = (d-x)^2 \]

### Step 8: Expand and Rearrange
Expanding the right side:
\[ 2x^2 = d^2 - 2dx + x^2 \]
Rearranging gives:
\[ x^2 + 2dx - d^2 = 0 \]

### Step 9: Solve the Quadratic Equation
Using the quadratic formula:
\[ x = \frac{-2d \pm \sqrt{(2d)^2 + 4d^2}}{2} \]
\[ x = \frac{-2d \pm 2d\sqrt{2}}{2} = -d + d\sqrt{2} \]
Thus:
\[ x = d(\sqrt{2} - 1) \]

### Final Position of Charge \( Q \)
Charge \( Q \) should be placed at a distance \( d(\sqrt{2} - 1) \) from charge \( q \).