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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 coulomb situated at point (1, 1, 1) is:

1. −(6i + 5j + 2k)
2. −(2i + 3j + k)
3. −(6i + 9j + k)
4. −(3i + 5j + 3k)

Correct answer: −(6i + 5j + 2k)

Solution

The electric field is the negative gradient of the potential, E = -∇V. Calculating partial derivatives of V(x, y, z) = 6x − 8xy − 8y + 6yz at (1, 1, 1), we get E = −(6i + 5j + 2k). The force on a charge q is F = qE, so for q = 2 C, F = −(6i + 5j + 2k).

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