Step-by-step Solution:

1. Identify the configuration and the task. We are given a system of two point charges: a positive charge +q at (0, 0, a/2) and a negative charge -q at (0, 0, -a/2). This configuration forms an electric dipole centered at the origin with its axis along the z-axis. We need to find the work done by the electric field when another positive point charge, let's call it $q_0$, is moved from an initial position A = (-a, 0, 0) to a final position B = (0, a, 0).

2. Recall the formula for work done by an electric field. The electrostatic field is a conservative field. The work done by the electric field in moving a charge $q_0$ from a point A to a point B is independent of the path taken and is given by the change in potential energy: $W = -\Delta U = -q_0(V_B - V_A) = q_0(V_A - V_B)$ where $V_A$ and $V_B$ are the electric potentials at points A and B, respectively.

3. Calculate the electric potential at the initial point A = (-a, 0, 0). The potential at point A is the algebraic sum of the potentials due to the two charges +q and -q. The potential due to a point charge Q at a distance r is V = kQ/r, where $k = 1/(4\pi\epsilon_0)$.

   • Distance of point A from the charge +q at $P_1(0, 0, a/2)$ is: $r_{1A} = \sqrt{(-a-0)^2 + (0-0)^2 + (0-a/2)^2} = \sqrt{a^2 + \frac{a^2}{4}} = \sqrt{\frac{5a^2}{4}} = \frac{a\sqrt{5}}{2}$
   • Distance of point A from the charge -q at $P_2(0, 0, -a/2)$ is: $r_{2A} = \sqrt{(-a-0)^2 + (0-0)^2 + (0-(-a/2))^2} = \sqrt{a^2 + \frac{a^2}{4}} = \sqrt{\frac{5a^2}{4}} = \frac{a\sqrt{5}}{2}$

   The total potential at A is: $V_A = V_{A,+q} + V_{A,-q} = \frac{k(+q)}{r_{1A}} + \frac{k(-q)}{r_{2A}}$ Since $r_{1A} = r_{2A}$, we have: $V_A = k q \left(\frac{1}{r_{1A}} - \frac{1}{r_{2A}}\right) = 0$

4. Calculate the electric potential at the final point B = (0, a, 0). Similarly, we calculate the potential at point B.

   • Distance of point B from the charge +q at $P_1(0, 0, a/2)$ is: $r_{1B} = \sqrt{(0-0)^2 + (a-0)^2 + (0-a/2)^2} = \sqrt{a^2 + \frac{a^2}{4}} = \sqrt{\frac{5a^2}{4}} = \frac{a\sqrt{5}}{2}$
   • Distance of point B from the charge -q at $P_2(0, 0, -a/2)$ is: $r_{2B} = \sqrt{(0-0)^2 + (a-0)^2 + (0-(-a/2))^2} = \sqrt{a^2 + \frac{a^2}{4}} = \sqrt{\frac{5a^2}{4}} = \frac{a\sqrt{5}}{2}$

   The total potential at B is: $V_B = V_{B,+q} + V_{B,-q} = \frac{k(+q)}{r_{1B}} + \frac{k(-q)}{r_{2B}}$ Since $r_{1B} = r_{2B}$, we have: $V_B = k q \left(\frac{1}{r_{1B}} - \frac{1}{r_{2B}}\right) = 0$

5. Alternative approach using the concept of equipotential surfaces. The given charge configuration is an electric dipole along the z-axis. The plane that is perpendicular to the dipole axis and passes through its center is called the equatorial plane. For this dipole, the equatorial plane is the xy-plane (the plane where z=0). Any point on the equatorial plane is equidistant from the two charges of the dipole. Therefore, the electric potential at any point on the equatorial plane is zero. $V_{eq} = \frac{kq}{r} + \frac{k(-q)}{r} = 0$ The initial point A(-a, 0, 0) and the final point B(0, a, 0) both have a z-coordinate of 0. This means both points lie on the equatorial plane of the dipole. Therefore, the potential at both points is zero: $V_A = 0$ and $V_B = 0$.

6. Calculate the work done. Using the formula from step 2: $W = q_0(V_A - V_B) = q_0(0 - 0) = 0$ The work done by the electric field is zero.

7. Conclusion. The work done is zero because the charge is moved between two points on the same equipotential surface. Option (D) is incorrect because the electrostatic force is conservative, making the work done path-independent.

Final Answer is C.