In a certain region of space, the potential is given by `V=k(2x^(2)-y^(2)+z^(2))`. The electric field at the point (1, 1, 1) has magnitude:

Options

A. (a)`ksqrt6`
B. (b)`2ksqrt6`
C. (c)`2ksqrt3`
D. (d)`4ksqrt3`

Explanation

To find the magnitude of the electric field at the point (1, 1, 1) given the potential \( V = k(2x^2 - y^2 + z^2) \), 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
The gradient in three dimensions is given by:
\[ \nabla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right) \]
Now, we need to compute the partial derivatives of \( V \).

1. **Partial derivative with respect to \( x \)**:
\[ \frac{\partial V}{\partial x} = k \cdot \frac{\partial}{\partial x}(2x^2 - y^2 + z^2) = k \cdot 4x \]

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

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

Thus, the gradient is:
\[ \nabla V = \left( 4kx, -2ky, 2kz \right) \]

### Step 3: Calculate the Electric Field
Now we can find the electric field:
\[ \mathbf{E} = -\nabla V = \left( -4kx, 2ky, -2kz \right) \]

### Step 4: Evaluate the Electric Field at the Point (1, 1, 1)
Substituting \( x = 1 \), \( y = 1 \), and \( z = 1 \):
\[ \mathbf{E} = \left( -4k(1), 2k(1), -2k(1) \right) = \left( -4k, 2k, -2k \right) \]

### Step 5: Calculate the Magnitude of the Electric Field
The magnitude of the electric field \( |\mathbf{E}| \) is given by:
\[ |\mathbf{E}| = \sqrt{(-4k)^2 + (2k)^2 + (-2k)^2} \]
Calculating this:
\[ |\mathbf{E}| = \sqrt{16k^2 + 4k^2 + 4k^2} = \sqrt{24k^2} = k\sqrt{24} = k \cdot 2\sqrt{6} \]

### Final Answer
The magnitude of the electric field at the point (1, 1, 1) is:
\[ |\mathbf{E}| = 2k\sqrt{6} \]

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