In a region, the potential is respresented by `V(x, y, z) = 6x - 8xy - 8y + 6yz`, where `V` is in volts and `x, y, z` are in meters. The electric force experienced by a charge of `2` coulomb situated at point `(1, 1, 1)` is

Options

A. `6sqrt(5)N`
B. `30N`
C. `24N`
D. `4sqrt(35)N`

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the Electric Field from the 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
Given the potential:
\[ V(x, y, z) = 6x - 8xy - 8y + 6yz \]
we need to compute the partial derivatives with respect to \(x\), \(y\), and \(z\).

1. **Partial derivative with respect to \(x\)**:
\[ \frac{\partial V}{\partial x} = 6 - 8y \]

2. **Partial derivative with respect to \(y\)**:
\[ \frac{\partial V}{\partial y} = -8x - 8 + 6z = -8x + 6z - 8 \]

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

### Step 3: Write the Electric Field Vector
Now substituting these derivatives into the electric field equation:
\[ \mathbf{E} = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right) \]
This gives:
\[ \mathbf{E} = -\left( (6 - 8y) \hat{i} + (-8x + 6z - 8) \hat{j} + (6y) \hat{k} \right) \]
Thus,
\[ \mathbf{E} = -(6 - 8y) \hat{i} + (8x - 6z + 8) \hat{j} - (6y) \hat{k} \]

### Step 4: Evaluate the Electric Field at Point (1, 1, 1)
Substituting \(x = 1\), \(y = 1\), and \(z = 1\):
\[ \mathbf{E} = -\left( 6 - 8(1) \right) \hat{i} + \left( 8(1) - 6(1) + 8 \right) \hat{j} - (6(1)) \hat{k} \]
Calculating each component:
1. **For \(\hat{i}\)**:
\[ -\left( 6 - 8 \right) = -(-2) = 2 \]
2. **For \(\hat{j}\)**:
\[ 8 - 6 + 8 = 10 \]
3. **For \(\hat{k}\)**:
\[ -6 \]
Thus, the electric field at point (1, 1, 1) is:
\[ \mathbf{E} = 2 \hat{i} + 10 \hat{j} - 6 \hat{k} \]

### Step 5: Calculate the Magnitude of the Electric Field
The magnitude of the electric field \(|\mathbf{E}|\) is given by:
\[ |\mathbf{E}| = \sqrt{(2)^2 + (10)^2 + (-6)^2} = \sqrt{4 + 100 + 36} = \sqrt{140} = 2\sqrt{35} \text{ N/C} \]

### Step 6: Calculate the Electric Force on the Charge
The electric force \(\mathbf{F}\) experienced by a charge \(Q\) in an electric field \(\mathbf{E}\) is given by:
\[ \mathbf{F} = Q \mathbf{E} \]
Given \(Q = 2 \text{ C}\):
\[ \mathbf{F} = 2 \cdot (2\sqrt{35}) = 4\sqrt{35} \text{ N} \]

### Final Answer
The electric force experienced by the charge of \(2\) coulombs situated at point \((1, 1, 1)\) is:
\[ \mathbf{F} = 4\sqrt{35} \text{ N} \]