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

Options

A. 8 along negative X-axis
B. 8 along positive X-axis
C. 16 along negative X-axis
D. 16 along positive Z-axis

Answer

To find the electric field at the point (1, 0, 2) 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 \( \vec{E} \) is related to the electric potential \( V \) by the equation:
\[ \vec{E} = -\nabla V \]
where \( \nabla V \) is the gradient of the potential.

### Step 2: Calculate the gradient of the potential
The gradient in three dimensions is given by:
\[ \nabla V = \left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right) \]
We need to compute the partial derivatives of \( V \) with respect to \( x \), \( y \), and \( z \).

### 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 \( y \)**:
\[ \frac{\partial V}{\partial y} = 0 \quad \text{(since V does not depend on y)} \]
3. **Partial derivative with respect to \( z \)**:
\[ \frac{\partial V}{\partial z} = 0 \quad \text{(since V does not depend on z)} \]

### Step 4: Write the gradient vector
Now, substituting the partial derivatives into the gradient:
\[ \nabla V = (8x) \hat{i} + 0 \hat{j} + 0 \hat{k} = 8x \hat{i} \]

### Step 5: Calculate the electric field
Now, substituting this into the electric field equation:
\[ \vec{E} = -\nabla V = -8x \hat{i} \]

### Step 6: Evaluate at the point (1, 0, 2)
Substituting \( x = 1 \):
\[ \vec{E} = -8(1) \hat{i} = -8 \hat{i} \]

### Final Answer
The electric field at the point (1, 0, 2) is:
\[ \vec{E} = -8 \hat{i} \text{ volts/meter} \]

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