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

Options

A. `+8` is x direction
B. 8 in -x direction
C. 16 in+x direction
D. 16 in -x direction

Explanation

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

### Step 1: Understand the relationship between electric potential and electric field
The electric field \( \vec{E} \) is related to the electric potential \( V \) by the equation:
\[ \vec{E} = -\nabla V \]
where \( \nabla V \) is the gradient of the potential \( V \).

### Step 2: Calculate the gradient of the potential
The potential given is:
\[ V = 4x^2 \]
We need to find the gradient \( \nabla V \). The gradient in Cartesian coordinates is given by:
\[ \nabla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right) \]

### Step 3: Compute the partial derivatives
1. **Partial derivative with respect to \( x \)**:
\[ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(4x^2) = 8x \]
2. **Partial derivative with respect to \( y \)**:
\[ \frac{\partial V}{\partial y} = 0 \quad (\text{since } V \text{ does not depend on } y) \]
3. **Partial derivative with respect to \( z \)**:
\[ \frac{\partial V}{\partial z} = 0 \quad (\text{since } V \text{ does not depend on } z) \]

Thus, the gradient is:
\[ \nabla V = (8x, 0, 0) \]

### Step 4: Substitute the coordinates into the gradient
At the point \( (1, 0, 2) \):
\[ \nabla V = (8 \cdot 1, 0, 0) = (8, 0, 0) \]

### Step 5: Calculate the electric field
Now, substituting into the equation for the electric field:
\[ \vec{E} = -\nabla V = -(8, 0, 0) = (-8, 0, 0) \]

### Step 6: Write the final answer in vector form
Thus, the electric field \( \vec{E} \) at the point \( (1, 0, 2) \) is:
\[ \vec{E} = -8 \hat{i} \, \text{(in volts/meter)} \]

### Summary of the Solution
The electric field \( E \) at the point \( (1, 0, 2) \) is:
\[ \vec{E} = -8 \hat{i} \, \text{V/m} \]

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