If potential (in volts) in a region is expressed as `V (x, y, z) = 6xy - y + 2yz`, the electric field (in `N//C)` at point `(1, 1, 0)` is

Options

A. `-(3hati _5hatj +3hatk)`
B. `-(6hati +5hatj +2hatk)`
C. `-(2hati =3hatj +hatjk)`
D. `-(6hati +9hatj + hatk)`

Answer

To find the electric field at the point (1, 1, 0) given the potential function \( V(x, y, z) = 6xy - y + 2yz \), we will follow these steps:

### Step 1: Understand the relationship between electric field and potential
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 gradient in three dimensions 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 \( x \)**:
\[ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(6xy - y + 2yz) = 6y \]

2. **Partial derivative with respect to \( y \)**:
\[ \frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(6xy - y + 2yz) = 6x - 1 + 2z \]

3. **Partial derivative with respect to \( z \)**:
\[ \frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(6xy - y + 2yz) = 2y \]

### Step 4: Write the gradient vector
Now we can write the gradient vector:
\[ \nabla V = \left( 6y, 6x - 1 + 2z, 2y \right) \]

### Step 5: Substitute the point (1, 1, 0)
Substituting \( x = 1 \), \( y = 1 \), and \( z = 0 \):
\[ \nabla V(1, 1, 0) = \left( 6 \cdot 1, 6 \cdot 1 - 1 + 2 \cdot 0, 2 \cdot 1 \right) = \left( 6, 6 - 1 + 0, 2 \right) = \left( 6, 5, 2 \right) \]

### Step 6: Calculate the electric field
Using the relationship \( \mathbf{E} = -\nabla V \):
\[ \mathbf{E} = -\left( 6, 5, 2 \right) = \left( -6, -5, -2 \right) \]

### Final Result
Thus, the electric field at the point \( (1, 1, 0) \) is:
\[ \mathbf{E} = -6 \hat{i} - 5 \hat{j} - 2 \hat{k} \text{ N/C} \]