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

Options

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

Explanation

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

### Step 1: Understand the Configuration
We have two charges:
- Charge `-q` located at the origin `(0, 0, 0)`.
- Charge `+q/2` located at the point `(a, 0, 0)`.

We need to find a point `P` on the X-axis (let's denote its position as `x`) where the electric field due to both charges cancels out.

### Step 2: Write the Expression for Electric Fields
The electric field `E` due to a point charge is given by:
\[ E = \frac{k \cdot |Q|}{r^2} \]
where `k` is Coulomb's constant, `Q` is the charge, and `r` is the distance from the charge to the point where we are measuring the electric field.

1. **Electric Field due to Charge `-q` (E1)**:
- The distance from the charge `-q` to point `P` at position `x` is `x`.
- Therefore, the electric field `E1` at point `P` due to charge `-q` is:
\[ E_1 = \frac{k \cdot q}{x^2} \]
(The direction of this field is towards the charge since it is negative.)

2. **Electric Field due to Charge `+q/2` (E2)**:
- The distance from the charge `+q/2` to point `P` is `|x - a|`.
- Therefore, the electric field `E2` at point `P` due to charge `+q/2` is:
\[ E_2 = \frac{k \cdot \frac{q}{2}}{(x - a)^2} \]
(The direction of this field is away from the charge since it is positive.)

### Step 3: Set Up the Equation for Electric Field to Vanish
For the electric field to vanish at point `P`, the magnitudes of `E1` and `E2` must be equal:
\[ E_1 = E_2 \]
Substituting the expressions we derived:
\[ \frac{k \cdot q}{x^2} = \frac{k \cdot \frac{q}{2}}{(x - a)^2} \]

### Step 4: Simplify the Equation
We can cancel `k` and `q` (assuming `q` is not zero) from both sides:
\[ \frac{1}{x^2} = \frac{1/2}{(x - a)^2} \]
Cross-multiplying gives:
\[ 2(x - a)^2 = x^2 \]

### Step 5: Expand and Rearrange
Expanding the left side:
\[ 2(x^2 - 2ax + a^2) = x^2 \]
This simplifies to:
\[ 2x^2 - 4ax + 2a^2 = x^2 \]
Rearranging gives:
\[ x^2 - 4ax + 2a^2 = 0 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -4a, c = 2a^2 \):
\[ x = \frac{4a \pm \sqrt{(-4a)^2 - 4 \cdot 1 \cdot 2a^2}}{2 \cdot 1} \]
Calculating the discriminant:
\[ x = \frac{4a \pm \sqrt{16a^2 - 8a^2}}{2} \]
\[ x = \frac{4a \pm \sqrt{8a^2}}{2} \]
\[ x = \frac{4a \pm 2\sqrt{2}a}{2} \]
\[ x = 2a \pm \sqrt{2}a \]

### Step 7: Find the Valid Solution
The two potential solutions are:
1. \( x = (2 + \sqrt{2})a \)
2. \( x = (2 - \sqrt{2})a \)

Since we are looking for a point on the X-axis between the two charges, we take the solution:
\[ x = (2 - \sqrt{2})a \]

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