The electric potential at any point `x,y and z` in metres is given by `V = 3x^(2)`. The electric field at a point `(2,0,1)` is

Options

A. `12 " Vm"^(-1)`
B. `-6 " Vm"^(-1)`
C. `6 " Vm"^(-1)`
D. `-12 " Vm"^(-1)`

Answer

To find the electric field at the point (2, 0, 1) given the electric potential \( V = 3x^2 \), we can 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.

### Step 2: Calculate the gradient of the potential
The potential \( V \) is given as a function of \( x \):
\[ V = 3x^2 \]
To find the electric field, we need to compute the partial derivatives of \( V \) with respect to \( x \), \( y \), and \( z \).

- The partial derivative with respect to \( x \):
\[ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(3x^2) = 6x \]

- The partial derivative with respect to \( y \):
\[ \frac{\partial V}{\partial y} = 0 \quad \text{(since \( V \) does not depend on \( y \))} \]

- The 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 we can write the electric field vector:
\[ \mathbf{E} = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right) \]
Substituting the computed derivatives:
\[ \mathbf{E} = -\left( 6x \hat{i} + 0 \hat{j} + 0 \hat{k} \right) = -6x \hat{i} \]

### Step 4: Evaluate the electric field at the point (2, 0, 1)
Now we substitute \( x = 2 \) into the electric field expression:
\[ \mathbf{E} = -6(2) \hat{i} = -12 \hat{i} \]

### Final Answer
Thus, the electric field at the point (2, 0, 1) is:
\[ \mathbf{E} = -12 \hat{i} \, \text{V/m} \]

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