Electric potential is given by
`V = 6x - 8xy^(2) - 8y + 6yz - 4z^(2)`
Then electric force acting on `2C` point charge placed on origin will be

Options

A. `2 N`
B. `6 N`
C. `8 N`
D. `20 N`

Explanation

To find the electric force acting on a 2C point charge placed at the origin, we need to follow these steps:

### Step 1: Identify the electric potential function
The electric potential \( V \) is given by:
\[ V = 6x - 8xy^2 - 8y + 6yz - 4z^2 \]

### Step 2: Calculate the electric field components
The electric field \( \vec{E} \) is related to the electric potential \( V \) by the negative gradient:
\[ \vec{E} = -\nabla V \]
This means we need to calculate the partial derivatives of \( V \) with respect to \( x \), \( y \), and \( z \).

#### Step 2.1: Calculate \( E_x \)
\[ E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x}(6x - 8xy^2 - 8y + 6yz - 4z^2) \]
Calculating the derivative:
\[ E_x = -6 \]

#### Step 2.2: Calculate \( E_y \)
\[ E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y}(6x - 8xy^2 - 8y + 6yz - 4z^2) \]
Calculating the derivative:
\[ E_y = -(-16y - 8 + 6z) = 16y + 8 - 6z \]
At the origin \( (0, 0, 0) \):
\[ E_y = 8 \]

#### Step 2.3: Calculate \( E_z \)
\[ E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(6x - 8xy^2 - 8y + 6yz - 4z^2) \]
Calculating the derivative:
\[ E_z = -(-8z + 6y) = 8z - 6y \]
At the origin \( (0, 0, 0) \):
\[ E_z = 0 \]

### Step 3: Combine the electric field components
The electric field at the origin is:
\[ \vec{E} = E_x \hat{i} + E_y \hat{j} + E_z \hat{k} = -6 \hat{i} + 8 \hat{j} + 0 \hat{k} \]

### Step 4: Calculate the magnitude of the electric field
The magnitude of the electric field \( E \) is given by:
\[ E = \sqrt{E_x^2 + E_y^2 + E_z^2} = \sqrt{(-6)^2 + (8)^2 + (0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{N/C} \]

### Step 5: Calculate the electric force on the charge
The electric force \( \vec{F} \) acting on a charge \( q \) in an electric field \( \vec{E} \) is given by:
\[ \vec{F} = q \vec{E} \]
For a charge \( q = 2C \):
\[ \vec{F} = 2 \, \text{C} \cdot ( -6 \hat{i} + 8 \hat{j} + 0 \hat{k}) = -12 \hat{i} + 16 \hat{j} \, \text{N} \]

### Step 6: Calculate the magnitude of the force
The magnitude of the force is:
\[ F = \sqrt{(-12)^2 + (16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \, \text{N} \]

### Final Answer
The electric force acting on the 2C point charge placed at the origin is:
\[ \vec{F} = 20 \, \text{N} \]

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