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 changes uniformly along the x-axis. Here’s the step-by-step solution:

### Step 1: Understand the relationship between electric potential and electric field
The electric field \( E \) is related to the electric potential \( V \) by the formula:
\[ E = -\frac{dV}{dx} \]
This means that the electric field is the negative gradient (or rate of change) of the electric potential.

### Step 2: Identify the change in electric potential
From the problem, we know that the electric potential decreases from 60 V to 20 V as we move from \( x = -2 \, m \) to \( x = +2 \, m \). Therefore, we can calculate the change in potential \( \Delta V \):
\[ \Delta V = V_{\text{final}} - V_{\text{initial}} = 20 \, V - 60 \, V = -40 \, V \]

### Step 3: Calculate the change in position
The change in position \( \Delta x \) over which this potential change occurs is:
\[ \Delta x = x_{\text{final}} - x_{\text{initial}} = 2 \, m - (-2 \, m) = 2 \, m + 2 \, m = 4 \, m \]

### Step 4: Substitute into the electric field formula
Now we can substitute \( \Delta V \) and \( \Delta x \) into the electric field equation:
\[ E = -\frac{\Delta V}{\Delta x} = -\frac{-40 \, V}{4 \, m} = \frac{40 \, V}{4 \, m} = 10 \, V/m \]

### Step 5: Determine the magnitude of the electric field
Since we are interested in the magnitude of the electric field, we take the absolute value:
\[ |E| = 10 \, V/m \]

### Conclusion
The magnitude of the electric field at the origin is \( 10 \, V/m \).

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