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 to find the electric field, we need to differentiate the potential function with respect to \( x \).

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

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

### Step 4: Evaluate the electric field at the point (-4, 0)
We need to find the electric field at the point \( (-4, 0) \). We substitute \( x = -4 \) into the equation for \( E \):
\[ E = -16(-4) = 64 \, \text{V/m} \]

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