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

Options

A. (a) The electric field E at all points on the x-axis has the same direction
B. (b) Electric field at all points on y-axis is along x-axis
C. (c) Work has to be done in bringing a test charge from `oo` to the origin
D. (d) The dipole moment is `2qd` along the x-axis

Explanation

To solve the problem involving two point charges \( +q \) and \( -q \) located at coordinates \( (-d, 0) \) and \( (d, 0) \) respectively, we will analyze the dipole moment, electric field, and work done in bringing a test charge from infinity to the origin.

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions**:
- The positive charge \( +q \) is located at \( (-d, 0) \).
- The negative charge \( -q \) is located at \( (d, 0) \).

2. **Calculate the Dipole Moment**:
- The dipole moment \( \vec{p} \) is given by the formula:
\[ \vec{p} = q \cdot \vec{d} \]
- The distance \( \vec{d} \) between the charges is \( 2d \) (from \( -d \) to \( d \)).
- Therefore, the dipole moment is:
\[ \vec{p} = q \cdot (2d) \hat{i} = 2qd \hat{i} \]
- The direction of the dipole moment is from the negative charge to the positive charge, which is along the negative x-axis.

3. **Determine the Electric Field Along the X-Axis**:
- The electric field \( \vec{E} \) at any point on the x-axis due to the charges can be calculated using the principle of superposition.
- For a point \( P \) on the x-axis, the electric field due to \( +q \) is directed away from the charge, while the field due to \( -q \) is directed towards the charge.
- The net electric field will depend on the position of point \( P \) relative to the charges.

4. **Analyze the Electric Field on the Y-Axis**:
- At any point on the y-axis, the electric field contributions from both charges will have components that result in a net electric field along the x-axis.
- This means that at all points on the y-axis, the resultant electric field will be directed along the x-axis.

5. **Calculate Work Done in Bringing a Test Charge from Infinity to the Origin**:
- The work done \( W \) in bringing a test charge \( q_0 \) from infinity to a point in an electric field is given by:
\[ W = q_0 (V_{\infty} - V_{0}) \]
- The potential \( V \) at a point due to a point charge is given by:
\[ V = \frac{kq}{r} \]
- At the origin (0,0), the contributions from both charges cancel each other out, resulting in a net potential of zero:
\[ V_{0} = 0 \]
- At infinity, the potential is also zero:
\[ V_{\infty} = 0 \]
- Therefore, the work done is:
\[ W = q_0 (0 - 0) = 0 \]

### Conclusion:
- The dipole moment is directed along the negative x-axis.
- The electric field at all points on the y-axis is directed along the x-axis.
- The work done in bringing a test charge from infinity to the origin is zero.