In a certain region of space, variation of potential with distance from origin as we move along x-axis is given by `V = 8 x^(2) + 2`, where x is the x-coordinate of a point in space. The magnitude of electric field at a point ( -4,0) is

Options

A. ` - 16 ` V/m
B. 16 V/m
C. ` -64` V/m
D. 64 V/m

Answer

To find the magnitude of the electric field at the point (-4, 0) given the potential function \( V = 8x^2 + 2 \), we can follow these steps:

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

### Step 2: Differentiate the potential function
Given the potential function:
\[ V = 8x^2 + 2 \]
we need to differentiate this with respect to \( x \):
\[ \frac{dV}{dx} = \frac{d}{dx}(8x^2 + 2) \]
Using the power rule for differentiation:
\[ \frac{dV}{dx} = 16x + 0 = 16x \]

### Step 3: Calculate the electric field
Now, substituting the derivative into the equation for the electric field:
\[ E = -\frac{dV}{dx} = -16x \]

### Step 4: Substitute the x-coordinate of the point
We need to find the electric field at the point (-4, 0). Thus, we substitute \( x = -4 \):
\[ E = -16(-4) = 64 \, \text{V/m} \]

### Step 5: State the magnitude of the electric field
The magnitude of the electric field at the point (-4, 0) is:
\[ |E| = 64 \, \text{V/m} \]

### Final Answer
The magnitude of the electric field at the point (-4, 0) is **64 V/m**.

---