Two point charges `(+Q)` and `(-2Q)` are fixed on the X-axis at positions `a` and `2a` from origin respectively. At what positions on the axis, the resultant electric field is zero

Options

A. Only `x=sqrt(2)a`
B. Only `x=-sqrt(2)a`
C. Both `x=+-sqrt(2)a`
D. `x=(3a)/(2)` only

Answer

To find the positions on the X-axis where the resultant electric field is zero due to the point charges \( +Q \) and \( -2Q \) fixed at positions \( a \) and \( 2a \) respectively, we can follow these steps:

### Step 1: Understand the Setup
We have two charges:
- Charge \( +Q \) at position \( a \)
- Charge \( -2Q \) at position \( 2a \)

We need to find the points on the X-axis where the electric field due to these charges cancels each other out, resulting in a net electric field of zero.

### Step 2: Identify Regions for Analysis
There are three regions to consider:
1. To the left of charge \( +Q \) (i.e., \( x < a \))
2. Between the charges \( +Q \) and \( -2Q \) (i.e., \( a < x < 2a \))
3. To the right of charge \( -2Q \) (i.e., \( x > 2a \))

### Step 3: Analyze Each Region
1. **Left of \( +Q \) (Region 1)**:
- Let’s denote a point \( x \) in this region. The electric field due to \( +Q \) points away from the charge (to the right), and the electric field due to \( -2Q \) points towards the charge (to the left).
- Therefore, in this region, the electric fields due to both charges are in opposite directions. It is possible for them to cancel each other out.

2. **Between \( +Q \) and \( -2Q \) (Region 2)**:
- In this region, the electric field due to \( +Q \) points to the right, while the electric field due to \( -2Q \) also points to the right.
- Thus, the net electric field cannot be zero in this region.

3. **Right of \( -2Q \) (Region 3)**:
- Let’s denote a point \( x \) in this region. The electric field due to \( +Q \) points to the right, and the electric field due to \( -2Q \) points to the left.
- Similar to Region 1, the electric fields are in opposite directions, and it is possible for them to cancel each other out.

### Step 4: Set Up the Electric Field Equations
For Region 1 (left of \( +Q \)):
- Let the point be at a distance \( x \) from the origin (where \( x < a \)).
- The distance from \( +Q \) is \( a - x \).
- The distance from \( -2Q \) is \( 2a - x \).

The electric field \( E \) at point \( x \) is given by:
\[ E = E_{+Q} + E_{-2Q} = \frac{kQ}{(a - x)^2} - \frac{2kQ}{(2a - x)^2} \]
Setting \( E = 0 \):
\[ \frac{kQ}{(a - x)^2} = \frac{2kQ}{(2a - x)^2} \]
Cancelling \( kQ \) (assuming \( Q \neq 0 \)):
\[ \frac{1}{(a - x)^2} = \frac{2}{(2a - x)^2} \]

### Step 5: Cross Multiply and Simplify
Cross-multiplying gives:
\[ (2a - x)^2 = 2(a - x)^2 \]
Expanding both sides:
\[ 4a^2 - 4ax + x^2 = 2(a^2 - 2ax + x^2) \]
\[ 4a^2 - 4ax + x^2 = 2a^2 - 4ax + 2x^2 \]
Rearranging:
\[ 4a^2 - 2a^2 = 2x^2 - x^2 \]
\[ 2a^2 = x^2 \]
Thus,
\[ x = \pm \sqrt{2}a \]

### Step 6: Determine Valid Solutions
- Since we are looking for a point left of \( +Q \), we take \( x = -\sqrt{2}a \).

### Final Answer
The position on the X-axis where the resultant electric field is zero is:
\[ x = -\sqrt{2}a \]