The electrical potential V at any point x, y, z (all in metres) in space is given by `V = 8x^(2) V` The electric field (in V/m) at points (1m, 0,2m) is given by

Options

A. `8i`
B. `16i`
C. `-16i`
D. `-8i`

Explanation

To find the electric field at the point (1m, 0, 2m) given the electric potential \( V = 8x^2 \) V, we will 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:
\[ \mathbf{E} = -\nabla V \]
where \( \nabla V \) is the gradient of the potential \( V \).

### Step 2: Calculate the gradient of the potential
The potential \( V \) is given as:
\[ V = 8x^2 \]
To find the electric field, we need to calculate the partial derivatives of \( V \) with respect to \( x \), \( y \), and \( z \).

#### Step 2.1: Calculate \( \frac{\partial V}{\partial x} \)
\[ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(8x^2) = 16x \]

#### Step 2.2: Calculate \( \frac{\partial V}{\partial y} \)
Since \( V \) does not depend on \( y \):
\[ \frac{\partial V}{\partial y} = 0 \]

#### Step 2.3: Calculate \( \frac{\partial V}{\partial z} \)
Since \( V \) does not depend on \( z \):
\[ \frac{\partial V}{\partial z} = 0 \]

### Step 3: Write the components of the electric field
Using the results from the partial derivatives, we can express the components of the electric field:
\[ E_x = -\frac{\partial V}{\partial x} = -16x \]
\[ E_y = -\frac{\partial V}{\partial y} = 0 \]
\[ E_z = -\frac{\partial V}{\partial z} = 0 \]

### Step 4: Evaluate the electric field at the point (1m, 0, 2m)
Now, we will substitute \( x = 1 \) m into the expression for \( E_x \):
\[ E_x = -16(1) = -16 \, \text{V/m} \]
Since \( E_y \) and \( E_z \) are both zero:
\[ \mathbf{E} = -16 \hat{i} + 0 \hat{j} + 0 \hat{k} = -16 \hat{i} \, \text{V/m} \]

### Final Answer
The electric field at the point (1m, 0, 2m) is:
\[ \mathbf{E} = -16 \hat{i} \, \text{V/m} \]