Relationship between Field and Potential: The electric field $\mathbf{E}$ is the negative gradient of the potential $V$ [1]. $\mathbf{E} = -\nabla V = -\left( \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k} \right)$

Partial Derivatives Calculation: Given $V = 6x - 8xy - 8y + 6yz$ $\frac{\partial V}{\partial x} = 6 - 8y$ $\frac{\partial V}{\partial y} = -8x - 8 + 6z$ $\frac{\partial V}{\partial z} = 6y$

Electric Field at Point (1, 1, 1): Substitute $x=1, y=1, z=1$: $E_x = -(6 - 8(1)) = -(-2) = 2$ $E_y = -(-8(1) - 8 + 6(1)) = -(-10) = 10$ $E_z = -(6(1)) = -6$ $\mathbf{E} = 2\hat{i} + 10\hat{j} - 6\hat{k} \text{ V/m}$

Electric Force Calculation: The force experienced by a charge $q$ in an electric field is $\mathbf{F} = q\mathbf{E}$ [2]. Given $q = 2 \text{ C}$: $\mathbf{F} = 2(2\hat{i} + 10\hat{j} - 6\hat{k}) = 4\hat{i} + 20\hat{j} - 12\hat{k} \text{ N}$

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