Rankers Physics

Electric Potential: Practice Problem & Solution

In a region, the potential is represented by $V(x, y, z) = 6x - 8xy - 8y + 6yz$, where $V$ is in volts and $x, y, z$ are in meters. The electric force experienced by a charge of $2 \text{ coulomb}$ situated at point $(1, 1, 1)$ is: (2014)
$6\sqrt{5} \text{ N}$
$30 \text{ N}$
$24 \text{ N}$
$4\sqrt{35} \text{ N}$

Solution Explained:

To solve this problem, we apply the core principles of Electric Potential. Understanding the underlying formula is key to arriving at the correct answer below:

$E = -\nabla V$. Partial derivatives at $(1,1,1)$: $E_x = -(6 - 8y) = 2$, $E_y = -(-8x - 8 + 6z) = 10$, $E_z = -(6y) = -6$. Magnitude $|E| = \sqrt{2^2 + 10^2 + (-6)^2} = \sqrt{140} = 2\sqrt{35} \text{ N/C}$. Force $F = qE = 2 \times 2\sqrt{35} = 4\sqrt{35} \text{ N}$.

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