To solve the problem, we need to calculate the work done by the electric field when moving a positive point charge from the point \((-a, 0, 0)\) to the point \((0, a, 0)\) in the presence of a positive charge at \((0, 0, a/2)\) and a negative charge at \((0, 0, -a/2)\).
### Step-by-Step Solution:
1. **Identify the Charges and Their Positions:**
- Let \( +Q \) be the positive charge located at \( (0, 0, a/2) \).
- Let \( -Q \) be the negative charge located at \( (0, 0, -a/2) \).
- We are moving a positive charge \( +q \) from the point \( (-a, 0, 0) \) to \( (0, a, 0) \).
2. **Understand the Electric Potential:**
- The electric potential \( V \) at a point in space due to a point charge is given by:
\[
V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}
\]
where \( r \) is the distance from the charge to the point where the potential is being calculated.
3. **Calculate the Electric Potential at the Initial and Final Points:**
- **Initial Point \((-a, 0, 0)\):**
- Distance from \( +Q \):
\[
r_1 = \sqrt{(-a)^2 + 0^2 + (0 - a/2)^2} = \sqrt{a^2 + (a/2)^2} = \sqrt{a^2 + \frac{a^2}{4}} = \sqrt{\frac{5a^2}{4}} = \frac{a\sqrt{5}}{2}
\]
- Distance from \( -Q \):
\[
r_2 = \sqrt{(-a)^2 + 0^2 + (0 + a/2)^2} = \sqrt{a^2 + (a/2)^2} = \frac{a\sqrt{5}}{2}
\]
- Total potential at \((-a, 0, 0)\):
\[
V_i = \frac{1}{4\pi\epsilon_0} \left( \frac{Q}{r_1} - \frac{Q}{r_2} \right) = \frac{1}{4\pi\epsilon_0} \left( \frac{Q}{\frac{a\sqrt{5}}{2}} - \frac{-Q}{\frac{a\sqrt{5}}{2}} \right) = 0
\]
- **Final Point \((0, a, 0)\):**
- Distance from \( +Q \):
\[
r_3 = \sqrt{(0)^2 + (a)^2 + (0 - a/2)^2} = \sqrt{a^2 + (a/2)^2} = \frac{a\sqrt{5}}{2}
\]
- Distance from \( -Q \):
\[
r_4 = \sqrt{(0)^2 + (a)^2 + (0 + a/2)^2} = \sqrt{a^2 + (a/2)^2} = \frac{a\sqrt{5}}{2}
\]
- Total potential at \((0, a, 0)\):
\[
V_f = \frac{1}{4\pi\epsilon_0} \left( \frac{Q}{r_3} - \frac{Q}{r_4} \right) = 0
\]
4. **Calculate the Work Done:**
- The work done \( W \) in moving a charge in an electric field is given by:
\[
W = q(V_f - V_i)
\]
- Since both \( V_i \) and \( V_f \) are equal to 0:
\[
W = q(0 - 0) = 0
\]
### Final Answer:
The work done by the electric field when moving the positive point charge from \((-a, 0, 0)\) to \((0, a, 0)\) is \( \boxed{0} \).
