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Electromagnetic theory miscellaneous

Electromagnetic Theory

  1. Given the potential function in free space to be V (x) = (50x2 + 50y2 + 50z2) volts, the magnitude (in volts/metre) and the direction of the electric field at a point (1, – 1, 1), where the dimensions are in metres, are—
    1. 100; (^ i + ^ j + ^ k)
    2. 100/√3; (^ i – ^ j + ^ k)
    3. 100 √3; [(–^ i – ^ j – ^ k)/√3]
    4. 100√3; [(–^ i – ^ j – ^ k)/3]
Correct Option: C


^^^
E = −
∂Vx
i +
∂Vy
j +
∂Vz
k volts/m
∂x∂y∂z

= −[100xi^ + 100yj^ + 100zk^]volts/m
At (1, – 1, 1) E = – 100 [^i – ^j + ^k] volts/m
Its direction is –^i + ^j – ^k/√3
and magnitude = 100√3



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