The potential of the electric field produced by point charge at any point (x, y, z) is given by `V = 3x^(2) + 5` , where x ,y are in are in metres and V is in volts. The intensity of the electric field at (-2,1,0) is :

Options

A. `+ 17 Vm^(-1)`
B. `-17 V m^(-1)`
C. `+12 Vm^(-1)`
D. `-12 V m^(-1)`

Explanation

To find the intensity of the electric field at the point (-2, 1, 0) given the potential \( V = 3x^2 + 5 \), we can follow these steps:

### Step 1: Understand the Relationship Between Electric Field and Potential
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, which can be expressed in terms of partial derivatives:
\[ \vec{E} = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right) \]

### Step 2: Calculate the Partial Derivatives of the Potential
1. **Partial derivative with respect to \( x \)**:
\[ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(3x^2 + 5) = 6x \]

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 3: Write the Expression for the Electric Field
Now substituting the partial derivatives into the electric field equation:
\[ \vec{E} = -\left( 6x \hat{i} + 0 \hat{j} + 0 \hat{k} \right) = -6x \hat{i} \]

### Step 4: Evaluate the Electric Field at the Point (-2, 1, 0)
Substituting \( x = -2 \) into the electric field expression:
\[ \vec{E} = -6(-2) \hat{i} = 12 \hat{i} \]

### Step 5: Write the Final Answer
The intensity of the electric field at the point (-2, 1, 0) is:
\[ \vec{E} = 12 \hat{i} \quad \text{(12 volts per meter in the positive x-direction)} \]