The electric potential V at any point x,y,z (all in metre) in space is given by `V=4x^2` volt. The electric field at the point `(1m, 0, 2m)` is ……………`V/m`.

Options

A. `-8hatiV//m`
B. `8hatiV//m`
C. `-18hatiV//m`
D. `28hatiV//m`

Explanation

To find the electric field at the point (1m, 0, 2m) given the electric potential \( V = 4x^2 \) volts, we can follow these steps:

### Step 1: Understand the relationship between electric potential and electric field
The electric field \( \mathbf{E} \) is related to the electric potential \( V \) by the equation:
\[ \mathbf{E} = -\nabla V \]
In Cartesian coordinates, this can be expressed as:
\[ \mathbf{E} = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right) \]

### Step 2: Differentiate the potential with respect to \( x \)
Given \( V = 4x^2 \), we need to find the derivative of \( V \) with respect to \( x \):
\[ \frac{\partial V}{\partial x} = \frac{d}{dx}(4x^2) = 8x \]

### Step 3: Calculate the electric field
Now, substituting the derivative into the electric field equation, we have:
\[ E_x = -\frac{\partial V}{\partial x} = -8x \]

### Step 4: Substitute the coordinates of the point (1m, 0, 2m)
Now, we substitute \( x = 1 \) into the equation for the electric field:
\[ E_x = -8(1) = -8 \, \text{V/m} \]

### Step 5: Determine the components of the electric field
Since \( V \) does not depend on \( y \) or \( z \), the components of the electric field in the \( y \) and \( z \) directions are zero:
\[ E_y = 0 \quad \text{and} \quad E_z = 0 \]

### Step 6: Write the final expression for the electric field
Thus, the electric field at the point (1m, 0, 2m) is:
\[ \mathbf{E} = -8 \hat{i} + 0 \hat{j} + 0 \hat{k} = -8 \hat{i} \, \text{V/m} \]

### Final Answer
The electric field at the point (1m, 0, 2m) is \(-8 \, \text{V/m}\) in the negative x-direction.

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