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. `-(6hat(i) + 9hat(j) + hat(k))`
B. `-(3hat(i) + 5hat(j) + 3hat(k))`
C. `-(6hat(i) + 5hat(j) + 2hat(k))`
D. `-(2hat(i) + 3hat(j) + hat(k))`

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 formula:
\[ \mathbf{E} = -\nabla V \]
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)
Now, we substitute \( x = 1 \), \( y = 1 \), and \( z = 0 \) into the gradient:
1. For \( \frac{\partial V}{\partial x} \):
\[ \frac{\partial V}{\partial x} = 6(1) = 6 \]

2. For \( \frac{\partial V}{\partial y} \):
\[ \frac{\partial V}{\partial y} = 6(1) - 1 + 2(0) = 6 - 1 = 5 \]

3. For \( \frac{\partial V}{\partial z} \):
\[ \frac{\partial V}{\partial z} = 2(1) = 2 \]

### Step 6: Write the electric field vector
Now, substituting these values into the electric field equation:
\[ \mathbf{E} = -\nabla V = -\left( 6, 5, 2 \right) = (-6, -5, -2) \]

### Step 7: Conclusion
The electric field at the point (1, 1, 0) is:
\[ \mathbf{E} = -6 \hat{i} - 5 \hat{j} - 2 \hat{k} \text{ N/C} \]