Work Done by Electric Field Moving Charge on Equipotential Plane

To determine the work done by the electric field, we consider the configuration of the charges:

1. Source Charge Configuration:
   A positive charge $+q$ is located at position $\vec{r}_1 = \left(0, 0, \frac{a}{2}\right)$ and a negative charge $-q$ is located at position $\vec{r}_2 = \left(0, 0, -\frac{a}{2}\right)$. This forms an electric dipole along the $z$-axis with its center at the origin $(0, 0, 0)$.

2. Electric Potential on the $xy$-plane ($z = 0$):
   The electric potential $V(x, y, z)$ at any point $(x, y, z)$ in space due to these two charges is given by the superposition principle:
   $V(x, y, z) = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{\sqrt{x^2 + y^2 + \left(z - \frac{a}{2}\right)^2}} + \frac{-q}{\sqrt{x^2 + y^2 + \left(z + \frac{a}{2}\right)^2}} \right]$

   For any point lying in the $xy$-plane, we have $z = 0$. Substituting $z = 0$ into the expression:
   $V(x, y, 0) = \frac{q}{4\pi\varepsilon_0} \left[ \frac{1}{\sqrt{x^2 + y^2 + \frac{a^2}{4}}} - \frac{1}{\sqrt{x^2 + y^2 + \frac{a^2}{4}}} \right] = 0$

   Therefore, the entire $xy$-plane is an equipotential surface with an electric potential of $V = 0$.

3. Work Done by the Electric Field:
   The initial position is $P_i = (-a, 0, 0)$ and the final position is $P_f = (0, a, 0)$. Both points lie in the $xy$-plane ($z = 0$):
   $V(P_i) = 0 \quad \text{and} \quad V(P_f) = 0$

   Since the electrostatic force is a conservative force, the work done by the electric field $W_{\text{field}}$ on a test charge $q_0$ moving from $P_i$ to $P_f$ is independent of the path taken and depends solely on the potential difference between the two points:
   $W_{\text{field}} = - \Delta U = q_0 \left( V(P_i) - V(P_f) \right)$
   $W_{\text{field}} = q_0 (0 - 0) = 0$

Thus, the work done by the electric field is zero.

Correct Answer: C (zero)