An electron entering field normally with a velocity 4 × 10 7 m s - 1 travels a distance of 0.10 m in an electric field of intensity 3200 V m - 1 . What is the deviation from its path?

Question

An electron entering field normally with a velocity 4 × 10 7 m s - 1 travels a distance of 0.10 m in an electric field of intensity 3200 V m - 1 . What is the deviation from its path?, 1.76 m m, 17.6 m m, 176 m m, 0.176 m m

MARKS App · Accepted Answer

Time to cross 0.10 m , t = 0.10 4 × 10 7 s Distance deviated in this time due to electric field = 1 2 ( e E m ) . t 2 = 1 2 × 1.6 × 10 − 19 × 3200 9.1 × 10 − 31 × ( 0.1 4 × 10 7 ) 2 m = 01.76 mm

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