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:
\[ \vec{E} = -\nabla V \]
In one dimension, this simplifies to:
\[ E = -\frac{dV}{dx} \]

### Step 2: Differentiate the potential with respect to \( x \)
Given the potential \( V = 3x^2 + 5 \), we need to differentiate this with respect to \( x \):
\[ \frac{dV}{dx} = \frac{d}{dx}(3x^2 + 5) = 6x \]

### Step 3: Calculate the electric field
Now, we can find the electric field \( E \) using the relation derived in Step 1:
\[ E = -\frac{dV}{dx} = -6x \]

### Step 4: Substitute the value of \( x \)
We need to evaluate the electric field at the point (-2, 1, 0). Here, \( x = -2 \):
\[ E = -6(-2) = 12 \, \text{V/m} \]

### Step 5: State the final answer
Thus, the intensity of the electric field at the point (-2, 1, 0) is:
\[ \boxed{12 \, \text{V/m}} \]

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