The electric field $\mathbf{E}$ is related to the electric potential $V$ by the negative gradient relationship: $\mathbf{E} = -\nabla V = -(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k})$ [NCERT Class 12, Sec 2.6.1].

1. Calculate Electric Field Components: Given $V = 6x - 8xy - 8y + 6yz$ $E_x = -\frac{\partial V}{\partial x} = -(6 - 8y)$ $E_y = -\frac{\partial V}{\partial y} = -(-8x - 8 + 6z) = 8x + 8 - 6z$

• $E_z = -\frac{\partial V}{\partial z} = -(6y) = -6y$

2. Evaluate Field at point $(1, 1, 1)$: Substitute $x=1, y=1, z=1$: $E_x = -(6 - 8(1)) = 2$ $E_y = 8(1) + 8 - 6(1) = 10$

• $E_z = -6(1) = -6$ So, $\mathbf{E} = 2\hat{i} + 10\hat{j} - 6\hat{k} \text{ N/C}$.

3. Calculate Electric Force: The force on a charge $q$ is $\mathbf{F} = q\mathbf{E}$ [NCERT Class 12, Sec 1.7]. Given $q = 2 \text{ C}$: $\mathbf{F} = 2(2\hat{i} + 10\hat{j} - 6\hat{k}) = 4\hat{i} + 20\hat{j} - 12\hat{k} \text{ N}$.

4. Calculate Magnitude of Force: $|\mathbf{F}| = \sqrt{4^2 + 20^2 + (-12)^2} = \sqrt{16 + 400 + 144} = \sqrt{560}$ $|\mathbf{F}| = \sqrt{16 \times 35} = 4\sqrt{35} \text{ N}$.