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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—
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- 100; (^ i + ^ j + ^ k)
- 100/√3; (^ i – ^ j + ^ k)
- 100 √3; [(–^ i – ^ j – ^ k)/√3]
- 100√3; [(–^ i – ^ j – ^ k)/3]
Correct Option: C
→ | ^ | ^ | ^ | ||||||
E = − | ![]() | i + | j + | 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