The electric potential decreases unifromly from 120 V to 80 V as one moves on the x-axis from x=-1 cm to x=+1 cm. The electric field at the origin

Options

A. must be equal to `20 V cm^-1`
B. must be equal to `20 Vm^-1`
C. may be greater than `20 Vm^-1`
D. may be less than `20 V cm^-1`

Explanation

To solve the problem, we need to find the electric field at the origin given that the electric potential decreases uniformly from 120 V to 80 V as we move along the x-axis from x = -1 cm to x = +1 cm.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We know that the electric potential \( V \) decreases from 120 V to 80 V as we move from \( x = -1 \) cm to \( x = +1 \) cm.
- This indicates that there is a uniform electric field acting along the x-axis.

2. **Finding the Change in Potential**:
- The change in potential \( \Delta V \) can be calculated as:
\[ \Delta V = V_{\text{final}} - V_{\text{initial}} = 80 \, \text{V} - 120 \, \text{V} = -40 \, \text{V} \]

3. **Calculating the Distance**:
- The distance \( d \) over which this change occurs is:
\[ d = x_{\text{final}} - x_{\text{initial}} = 1 \, \text{cm} - (-1 \, \text{cm}) = 2 \, \text{cm} = 0.02 \, \text{m} \]

4. **Using the Relationship Between Electric Field and Potential**:
- The relationship between electric field \( E \) and potential difference \( \Delta V \) is given by:
\[ \Delta V = -E \cdot d \]
- Rearranging this gives:
\[ E = -\frac{\Delta V}{d} \]

5. **Substituting Values**:
- Substituting the values we found:
\[ E = -\frac{-40 \, \text{V}}{0.02 \, \text{m}} = \frac{40 \, \text{V}}{0.02 \, \text{m}} = 2000 \, \text{V/m} \]

6. **Direction of the Electric Field**:
- Since the potential decreases in the positive x-direction, the electric field must point in the direction of decreasing potential, which is towards the left (negative x-direction).
- Therefore, the electric field at the origin is:
\[ E = -2000 \, \text{V/m} \, \text{(in the negative x-direction)} \]

### Final Answer:
The electric field at the origin is \( 2000 \, \text{V/m} \) directed towards the negative x-axis.