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
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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Problem 2.
Transcribed Image Text: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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