The electric potential decreases uniformly from 180V to 20V as one moves on the X-axis from `x=-2cm ` to `x= +2cm`. The electric field at the origin:

Options

A. must be equal to 40 V/cm
B. may be equal to 40 V/cm
C. may be less than 30 V/m
D. may be less than 40 V/cm

Explanation

To solve the problem, we need to determine the electric field at the origin based on the given electric potential values and their corresponding positions on the x-axis.

### Step-by-Step Solution:

1. **Identify the Change in Electric Potential (ΔV)**:
The electric potential decreases from 180V to 20V as we move from `x = -2 cm` to `x = +2 cm`.
\[ \Delta V = V_{\text{final}} - V_{\text{initial}} = 20V - 180V = -160V \]

2. **Identify the Change in Distance (Δx)**:
The distance over which this change occurs is from `-2 cm` to `+2 cm`.
\[ \Delta x = x_{\text{final}} - x_{\text{initial}} = 2 cm - (-2 cm) = 4 cm \]

3. **Calculate the Electric Field (E)**:
The electric field is defined as the negative gradient of the electric potential. In one dimension, it can be calculated using the formula:
\[ E = -\frac{\Delta V}{\Delta x} \]
Substituting the values we found:
\[ E = -\frac{-160V}{4 cm} = \frac{160V}{4 cm} = 40 \, \text{V/cm} \]

4. **Determine the Direction of the Electric Field**:
Since the potential decreases from 180V to 20V, the electric field points in the direction of decreasing potential. Thus, the electric field at the origin (x = 0) points from the higher potential (180V) towards the lower potential (20V), which is in the negative x-direction.

5. **Conclusion**:
The electric field at the origin is:
\[ E = 40 \, \text{V/cm} \, \text{(in the negative x-direction)} \]