The electric potential decreases uniformly from V to -V along X-axis in a coordinate system as we moves from a point `(-x_(0), 0)` to `(x_(0), 0)`, then the electric field at the origin:

Explanation

To find the electric field at the origin when the electric potential decreases uniformly from \( V \) to \( -V \) along the x-axis, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
The electric potential \( V \) decreases uniformly from \( V \) at the point \( (-x_0, 0) \) to \( -V \) at the point \( (x_0, 0) \). We need to determine the electric field at the origin \( (0, 0) \).

2. **Identify the Change in Potential**:
The potential at the left point \( (-x_0, 0) \) is \( V \) and at the right point \( (x_0, 0) \) is \( -V \). The total change in potential \( \Delta V \) as we move from \( -x_0 \) to \( x_0 \) is:
\[ \Delta V = V - (-V) = V + V = 2V \]

3. **Distance Over Which the Change Occurs**:
The distance over which this potential change occurs is from \( -x_0 \) to \( x_0 \), which is:
\[ d = x_0 - (-x_0) = 2x_0 \]

4. **Calculating the Electric Field**:
The electric field \( E \) is related to the change in electric potential \( \Delta V \) and the distance \( d \) over which this change occurs by the formula:
\[ E = -\frac{\Delta V}{d} \]
Substituting the values we found:
\[ E = -\frac{2V}{2x_0} = -\frac{V}{x_0} \]

5. **Direction of the Electric Field**:
Since the potential decreases from \( V \) to \( -V \), the electric field points in the direction of decreasing potential, which is towards the negative x-direction. Therefore, the electric field at the origin is:
\[ E = -\frac{V}{x_0} \hat{i} \]
where \( \hat{i} \) is the unit vector in the positive x-direction.

### Final Answer:
The electric field at the origin is given by:
\[ E = -\frac{V}{x_0} \hat{i} \]