If on the x - axis electric potential decreases uniform from 60 V to 20 V between x = -2 m to x = +2 m then the magnitude of electric field at the origin

Options

A. Must be 10 V/m
B. May be greater than 10 V/m
C. is zero
D. is 5 V/m

Answer

To solve the problem, we need to find the magnitude of the electric field at the origin given the electric potential decreases uniformly from 60 V to 20 V between x = -2 m and x = +2 m.

### Step-by-step Solution:

1. **Identify the given values:**
- Electric potential at \( x = -2 \, \text{m} \) (let's call this \( V_1 \)) = 60 V
- Electric potential at \( x = +2 \, \text{m} \) (let's call this \( V_2 \)) = 20 V
- The distance between these two points (let's call this \( \Delta r \)) = \( r_2 - r_1 = 2 - (-2) = 4 \, \text{m} \)

2. **Calculate the change in electric potential (\( \Delta V \)):**
\[ \Delta V = V_2 - V_1 = 20 \, \text{V} - 60 \, \text{V} = -40 \, \text{V} \]

3. **Calculate the electric field (\( E \)):**
The electric field is defined as the negative gradient of the electric potential:
\[ E = -\frac{\Delta V}{\Delta r} \]
Substitute the values:
\[ E = -\frac{-40 \, \text{V}}{4 \, \text{m}} = \frac{40 \, \text{V}}{4 \, \text{m}} = 10 \, \text{V/m} \]

4. **Determine the direction of the electric field:**
The negative sign in the change of potential indicates that the electric field points in the direction of decreasing potential. Since the potential is decreasing from 60 V to 20 V as we move from \( x = -2 \, \text{m} \) to \( x = +2 \, \text{m} \), the electric field at the origin (which is at \( x = 0 \)) will point towards the left (negative x-direction).

5. **State the final answer:**
The magnitude of the electric field at the origin is:
\[ |E| = 10 \, \text{V/m} \]

### Final Answer:
The magnitude of the electric field at the origin is \( 10 \, \text{V/m} \).