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Step-by-Step Solution
Step 1: Write down the given potential function
The electric potential is given as
$V(x, y, z) = 6x - 8xy - 8y + 6yz.$
Step 2: Recall the relation between electric field and potential
The electric field $ \vec{E} $ is related to the electric potential $V$ by
$$
\vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat{i} \;+\; \frac{\partial V}{\partial y}\hat{j} \;+\; \frac{\partial V}{\partial z}\hat{k}\right).
$$
Step 3: Compute partial derivatives of $V$
$ \displaystyle \frac{\partial V}{\partial x}
= \frac{\partial}{\partial x}(6x - 8xy - 8y + 6yz)
= 6 - 8y. $
$ \displaystyle \frac{\partial V}{\partial y}
= \frac{\partial}{\partial y}(6x - 8xy - 8y + 6yz)
= -8x - 8 + 6z. $
$ \displaystyle \frac{\partial V}{\partial z}
= \frac{\partial}{\partial z}(6x - 8xy - 8y + 6yz)
= 6y. $
Step 4: Substitute into the electric field expression
The electric field vector is then
$$
\vec{E} = -\big[(6 - 8y)\hat{i} + (-8x - 8 + 6z)\hat{j} + (6y)\hat{k}\big].
$$
Step 5: Evaluate the electric field at the point (1, 1, 1)
Substitute $x=1$, $y=1$, and $z=1$ into the partial derivatives:
$ \frac{\partial V}{\partial x}\bigg|_{(1,1,1)} = 6 - 8(1) = -2. $
$ \frac{\partial V}{\partial y}\bigg|_{(1,1,1)} = -8(1) - 8 + 6(1) = -8 - 8 + 6 = -10. $
$ \frac{\partial V}{\partial z}\bigg|_{(1,1,1)} = 6(1) = 6. $
Hence,
$$
\vec{E}\big|_{(1,1,1)}
= -\big[(-2)\hat{i} + (-10)\hat{j} + (6)\hat{k}\big]
= 2\hat{i} + 10\hat{j} - 6\hat{k}.
$$
Step 6: Find the magnitude of the electric field
Compute the magnitude:
$$
|\vec{E}| = \sqrt{(2)^2 + (10)^2 + (-6)^2}
= \sqrt{4 + 100 + 36}
= \sqrt{140}
= 2\sqrt{35}.
$$
Step 7: Calculate the force experienced by the charge
The force on a charge $q$ in an electric field $\vec{E}$ is given by $ \vec{F} = q \,\vec{E}.$ Here, $q = 2\, \text{C}.$ So,
$$
|\vec{F}| = q \,|\vec{E}|
= 2 \times 2\sqrt{35}
= 4\sqrt{35}\,\text{N}.
$$
Final Answer
The electric force experienced by the 2 C charge at (1, 1, 1) is
$4\sqrt{35}\,\text{N}.$
