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), we will use the relationship between electric field (E) and electric potential (V). The electric field is defined as the negative gradient of the electric potential:

\[ E = -\frac{dV}{dx} \]

Given the potential function:

\[ V = 8x^2 + 2 \]

we will follow these steps:

### Step 1: Differentiate the potential function with respect to x
We need to find the derivative of V with respect to x:

\[ \frac{dV}{dx} = \frac{d}{dx}(8x^2 + 2) \]

### Step 2: Calculate the derivative
Using the power rule of differentiation:

\[ \frac{dV}{dx} = 16x \]

### Step 3: Evaluate the derivative at x = -4
Now we will substitute \( x = -4 \) into the derivative to find the electric field at that point:

\[ \frac{dV}{dx} \bigg|_{x=-4} = 16(-4) = -64 \]

### Step 4: Calculate the electric field
Now, we can find the electric field using the relationship we established:

\[ E = -\frac{dV}{dx} = -(-64) = 64 \]

### Conclusion
The magnitude of the electric field at the point (-4, 0) is:

\[ E = 64 \, \text{N/C} \]