Solution:

To calculate the electric field in the overlap region, we use the principle of superposition of electric fields. Let's analyze:

Step 1: Electric field due to one sphere
For a uniformly charged non-conducting sphere, the electric field inside the sphere at a distance $\vec{r}$ from the center is:

$$\vec{E}_{\text{sphere}} = \frac{\rho}{3\epsilon_0} \vec{r}$$

Here:
- $\rho$ is the charge density of the sphere,
- $\epsilon_0$ is the permittivity of free space,
- $\vec{r}$ is the position vector from the center of the sphere.

Step 2: Contribution of both spheres in the overlap region
- For the positively charged sphere ($+\rho$), the electric field at any point in the overlap region is directed **away** from its center, proportional to $\vec{r}_1$ (distance from its center).
- For the negatively charged sphere ($-\rho$), the electric field at any point in the overlap region is directed **toward** its center, proportional to $\vec{r}_2$ (distance from its center).

Thus, the net electric field is the vector sum:

$$\vec{E}_{\text{net}} = \frac{\rho}{3\epsilon_0} \vec{r}_1 + \frac{-\rho}{3\epsilon_0} \vec{r}_2$$

Step 3: Relation between $\vec{r}_1$, $\vec{r}_2$, and $\vec{d}$
In the overlap region, $\vec{r}_1 - \vec{r}_2 = \vec{d}$, where $\vec{d}$ is the displacement vector between the centers of the two spheres.

Substitute this into the expression for $\vec{E}_{\text{net}}$:

$$\vec{E}_{\text{net}} = \frac{\rho}{3\epsilon_0} (\vec{r}_1 - \vec{r}_2) = \frac{\rho}{3\epsilon_0} \vec{d}$$

Final Answer:

The electric field in the overlap region is:

$${\frac{\rho}{3\epsilon_0} \vec{d}}$$