Two identical charge of value Q each are placed at (-a,0) and (a , 0) . The end coordinates of the point where the net electric field is zero and maximum are respectively-

Options

A. (0,0), (0,0)
B. `(0,a/sqrt2),(0,0)`
C. `(0,0),(0,a/sqrt2)`
D. `(a/sqrt2,0),(0,a/sqrt2)`

Answer

To solve the problem, we need to find the coordinates where the net electric field due to two identical charges \( Q \) placed at points \((-a, 0)\) and \((a, 0)\) is zero and where it is maximum.

### Step-by-step Solution:

1. **Understanding the Setup**:
We have two identical charges \( Q \) located at \((-a, 0)\) and \((a, 0)\). We need to analyze the electric field created by these charges at various points along the y-axis (vertical line).

2. **Electric Field Due to a Point Charge**:
The electric field \( E \) due to a point charge \( Q \) at a distance \( r \) is given by:
\[ E = \frac{kQ}{r^2} \]
where \( k \) is Coulomb's constant.

3. **Distance Calculation**:
For a point on the y-axis at \((0, y)\), the distance from each charge to this point is:
\[ r = \sqrt{a^2 + y^2} \]

4. **Electric Field Components**:
The electric field at point \((0, y)\) due to the charge at \((-a, 0)\) will point away from the charge, and similarly for the charge at \((a, 0)\). The vertical components of the electric fields from both charges will add up, while the horizontal components will cancel each other out.

5. **Net Electric Field Calculation**:
The net electric field \( E_{\text{net}} \) in the y-direction is given by:
\[ E_{\text{net}} = 2E \cos(\theta) \]
where \( \cos(\theta) = \frac{y}{\sqrt{a^2 + y^2}} \). Thus,
\[ E_{\text{net}} = 2 \cdot \frac{kQ}{a^2 + y^2} \cdot \frac{y}{\sqrt{a^2 + y^2}} = \frac{2kQy}{(a^2 + y^2)^{3/2}} \]

6. **Finding Points Where Electric Field is Zero**:
To find where the electric field is zero, set \( E_{\text{net}} = 0 \):
\[ \frac{2kQy}{(a^2 + y^2)^{3/2}} = 0 \]
This implies \( y = 0 \). Therefore, the net electric field is zero at the origin \((0, 0)\).

7. **Finding Maximum Electric Field**:
To find where the electric field is maximum, we differentiate \( E_{\text{net}} \) with respect to \( y \) and set the derivative to zero:
\[ \frac{dE_{\text{net}}}{dy} = 0 \]
After differentiating and simplifying, we find that \( y^2 = a^2/3 \). Thus,
\[ y = \pm \frac{a}{\sqrt{3}} \]
The coordinates where the electric field is maximum are \((0, \frac{a}{\sqrt{3}})\) and \((0, -\frac{a}{\sqrt{3}})\).

### Final Answer:
- The coordinates where the net electric field is zero: \((0, 0)\)
- The coordinates where the net electric field is maximum: \((0, \frac{a}{\sqrt{3}})\) and \((0, -\frac{a}{\sqrt{3}})\)