Two point charges +q and -q are held fixed at `(-a,0)` and `(a,0)` respectively of a x-y coordinate system then

Options

A. the electric field E at all point on the axis has the same direction.
B. E at all points on the y-axis is along `hati`
C. positive work is done in bringing a test charge from infinity to the origin
D. All of the above

Explanation

To solve the problem, we need to analyze the electric field created by two point charges, +q and -q, located at (-a, 0) and (a, 0) respectively.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two charges: +q at point (-a, 0) and -q at point (a, 0).
- The x-axis is the line along which the charges are placed, and the y-axis is perpendicular to it.

2. **Electric Field Due to Each Charge**:
- 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.

3. **Electric Field at the Origin (0, 0)**:
- At the origin, the electric field due to +q is directed away from the charge (to the right) and the electric field due to -q is directed towards the charge (to the left).
- Since both charges are equidistant from the origin, the magnitudes of the electric fields due to both charges at the origin are equal, but they are in opposite directions. Thus, they cancel each other out:
\[ E_{\text{net}} = E_{+q} + E_{-q} = 0 \]

4. **Electric Field Along the x-axis**:
- If we move along the x-axis (to the left or right of the origin), the electric field due to the positive charge (+q) will always point away from it (to the right), while the electric field due to the negative charge (-q) will always point towards it (to the left).
- As we move away from the origin towards the left, the influence of the positive charge decreases, and the influence of the negative charge increases. Thus, the net electric field will point to the left.
- Conversely, as we move to the right, the influence of the negative charge decreases, and the influence of the positive charge increases, making the net electric field point to the right.

5. **Electric Field Along the y-axis**:
- On the y-axis, the electric fields due to both charges will have components that cancel each other out. The y-components of the electric fields from both charges will be equal in magnitude but opposite in direction, thus resulting in a net electric field that is directed along the x-axis (i.e., along the i-cap direction).

6. **Potential at the Origin**:
- The electric potential (V) at the origin due to both charges can be calculated as:
\[ V = k \frac{+q}{a} + k \frac{-q}{a} = 0 \]
- Since the potential at infinity is also 0, the work done in bringing a test charge from infinity to the origin is:
\[ W = Q \Delta V = Q (0 - 0) = 0 \]

7. **Conclusion**:
- The electric field at all points on the x-axis does not have the same direction.
- The electric field at all points on the y-axis is directed along the i-cap direction.
- The work done in bringing a test charge from infinity to the origin is zero.

### Final Answer:
The correct option is B: The electric field at all points on the y-axis is along the i-cap direction.