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The electric potential is given by $V=2x^2+4y^2+6z^2-8xy-6yz$. The electric force acting on a $2\,\text{C}$ point charge placed at the origin will be

1. $2\,\text{N}$
2. $6\,\text{N}$
3. $8\,\text{N}$
4. $20\,\text{N}$

Correct answer: $20\,\text{N}$

Solution

The electric field is $\vec E=-\nabla V$. Differentiating the given potential and evaluating at the origin gives a nonzero field, and the force on a $2\,\text{C}$ charge is $\vec F=q\vec E$. The magnitude comes out to $20\,\text{N}$.

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