Two point charge `-q` and `+q//2` are situated at the origin and the point `(a,0,0)` respectively. The point along the `X`-axis where the electic field Vanishes is

Options

A. `x = (a)/(sqrt2)`
B. `x = sqrt2a`
C. `x = (sqrt(2a))/(sqrt2 - 1)`
D. `x = (sqrt(2a))/(sqrt2 + 1)`

Answer

To find the point along the X-axis where the electric field due to the two point charges \( -q \) (located at the origin) and \( +\frac{q}{2} \) (located at the point \( (a, 0, 0) \)) vanishes, we can follow these steps:

### Step 1: Understand the Electric Field Contributions
The electric field \( E \) due to a point charge is given by the formula:
\[ E = k \frac{|Q|}{r^2} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge to the point where the field is being calculated.

### Step 2: Set Up the Problem
Let’s denote the point where the electric field vanishes as \( x \) on the X-axis. The electric field at point \( x \) due to the charge \( -q \) at the origin (0, 0, 0) will be directed towards the charge (since it is negative), and the electric field due to the charge \( +\frac{q}{2} \) at \( (a, 0, 0) \) will be directed away from the charge (since it is positive).

### Step 3: Write the Electric Field Equations
1. The distance from the charge \( -q \) to the point \( x \) is \( |x| \).
2. The distance from the charge \( +\frac{q}{2} \) to the point \( x \) is \( |a - x| \).

The electric field at point \( x \) due to \( -q \) is:
\[ E_{-q} = k \frac{q}{x^2} \]
The electric field at point \( x \) due to \( +\frac{q}{2} \) is:
\[ E_{+\frac{q}{2}} = k \frac{\frac{q}{2}}{(a - x)^2} \]

### Step 4: Set the Electric Fields Equal
For the electric field to vanish, the magnitudes of these two electric fields must be equal:
\[ k \frac{q}{x^2} = k \frac{\frac{q}{2}}{(a - x)^2} \]

### Step 5: Simplify the Equation
We can cancel \( k \) and \( q \) (assuming \( q \neq 0 \)):
\[ \frac{1}{x^2} = \frac{1/2}{(a - x)^2} \]
Cross-multiplying gives:
\[ 2(a - x)^2 = x^2 \]

### Step 6: Expand and Rearrange
Expanding the left side:
\[ 2(a^2 - 2ax + x^2) = x^2 \]
This simplifies to:
\[ 2a^2 - 4ax + 2x^2 = x^2 \]
Rearranging gives:
\[ x^2 - 4ax + 2a^2 = 0 \]

### Step 7: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -4a, c = 2a^2 \):
\[ x = \frac{4a \pm \sqrt{(-4a)^2 - 4 \cdot 1 \cdot 2a^2}}{2 \cdot 1} \]
\[ x = \frac{4a \pm \sqrt{16a^2 - 8a^2}}{2} \]
\[ x = \frac{4a \pm \sqrt{8a^2}}{2} \]
\[ x = \frac{4a \pm 2\sqrt{2}a}{2} \]
\[ x = 2a \pm \sqrt{2}a \]

### Step 8: Determine the Valid Solution
The two possible solutions are:
1. \( x = (2 + \sqrt{2})a \)
2. \( x = (2 - \sqrt{2})a \)

Since we are looking for a point along the X-axis where the electric field vanishes, we need to consider the position relative to the charges. The point \( (2 - \sqrt{2})a \) is between the two charges and is valid.

### Final Answer
The point along the X-axis where the electric field vanishes is:
\[ x = (2 - \sqrt{2})a \]