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

Answer

To solve the problem, we need to find the electric force acting on a 2C point charge placed at the origin given the electric potential \( V = 6x - 8xy^2 - 8y + 6yz - 4z^2 \). We will follow these steps:

### Step 1: Calculate the Electric Field
The electric field \( \mathbf{E} \) is related to the electric potential \( V \) by the equation:
\[ \mathbf{E} = -\nabla V \]
This means we need to calculate the partial derivatives of \( V \) with respect to \( x \), \( y \), and \( z \).

### Step 2: Find \( E_x \)
Calculate the electric field in the x-direction:
\[ E_x = -\frac{\partial V}{\partial x} \]
Given \( V = 6x - 8xy^2 - 8y + 6yz - 4z^2 \):
- The derivative of \( 6x \) with respect to \( x \) is \( 6 \).
- The derivative of \( -8xy^2 \) with respect to \( x \) is \( -8y^2 \).
- The other terms are constants with respect to \( x \) and their derivatives are \( 0 \).

Thus,
\[ E_x = -\left(6 - 8y^2\right) = -6 + 8y^2 \]

### Step 3: Find \( E_y \)
Calculate the electric field in the y-direction:
\[ E_y = -\frac{\partial V}{\partial y} \]
Using the same potential \( V \):
- The derivative of \( 6x \) with respect to \( y \) is \( 0 \).
- The derivative of \( -8xy^2 \) with respect to \( y \) is \( -16xy \).
- The derivative of \( -8y \) is \( -8 \).
- The derivative of \( 6yz \) with respect to \( y \) is \( 6z \).
- The derivative of \( -4z^2 \) is \( 0 \).

Thus,
\[ E_y = -\left(-16xy - 8 + 6z\right) = 16xy + 8 - 6z \]

### Step 4: Find \( E_z \)
Calculate the electric field in the z-direction:
\[ E_z = -\frac{\partial V}{\partial z} \]
Again using the potential \( V \):
- The derivative of \( 6x \) with respect to \( z \) is \( 0 \).
- The derivative of \( -8xy^2 \) with respect to \( z \) is \( 0 \).
- The derivative of \( -8y \) with respect to \( z \) is \( 0 \).
- The derivative of \( 6yz \) with respect to \( z \) is \( 6y \).
- The derivative of \( -4z^2 \) is \( -8z \).

Thus,
\[ E_z = -\left(6y - 8z\right) = -6y + 8z \]

### Step 5: Evaluate Electric Field at the Origin
Now, we evaluate \( E_x \), \( E_y \), and \( E_z \) at the origin \((0, 0, 0)\):
- \( E_x(0, 0, 0) = -6 + 8(0)^2 = -6 \, \text{N/C} \)
- \( E_y(0, 0, 0) = 16(0)(0) + 8 - 6(0) = 8 \, \text{N/C} \)
- \( E_z(0, 0, 0) = -6(0) + 8(0) = 0 \, \text{N/C} \)

Thus, the electric field vector at the origin is:
\[ \mathbf{E} = -6 \hat{i} + 8 \hat{j} + 0 \hat{k} \]

### Step 6: 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 7: Calculate the Electric Force
The electric force \( \mathbf{F} \) acting on a charge \( Q \) in an electric field \( \mathbf{E} \) is given by:
\[ \mathbf{F} = Q \mathbf{E} \]
Given \( Q = 2 \, \text{C} \):
\[ \mathbf{F} = 2 \, \text{C} \cdot 10 \, \text{N/C} = 20 \, \text{N} \]

### Final Answer
The electric force acting on the 2C point charge placed at the origin is \( 20 \, \text{N} \).

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