The electric potential in a region of space is given by V(x, y, z) = A(xy + yz + zx) . Given that A is 10 SI units, determine the magnitude of the electric field at the point (1 m, 1 m, 1 m) .

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Correct answer: B. 20 3 N C^ -1

The electric field E is related to the electric potential V by the negative gradient of the potential: E = -( (∂ V)/(∂ x) i + (∂ V)/(∂ y) j + (∂ V)/(∂ z) k ) Given V(x, y, z) = A(xy + yz + zx) and A = 10 , we have: E_x = -(∂ V)/(∂ x) = -A(y + z) E_y = -(∂ V)/(∂ y) = -A(x + z) E_z = -(∂ V)/(∂ z) = -A(x + y) Substituting x = 1 , y = 1 , z = 1 and A = 10 : E_x = -10(1 + 1) = -20 E_y = -10(1 + 1) = -20 E_z = -10(1 + 1) = -20 The electric field vector is E = -20 i - 20 j - 20 k . The magnitude of the electric field is: | E | = (-20)^2 + (-20)^2 + (-20)^2 = 400 + 400 + 400 = 1200 = 20 3 N C^ -1

The electric potential in a region of space is given by V(x, y, z)...

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