Q27 / 250 · NTA Abhyas (NEET) 2020 · Physics · Electrostatics

An electron entering field normally with a velocity 4×107 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?

  • Correct answer
    A
    1.76 mm
  • B
    17.6 mm
  • C
    176 mm
  • D
    0.176 mm

Solution

Time to cross 0.10 m, t= 0.10 4× 10 7 s

Distance deviated in this time due to electric field = 1 2 ( eE m ).  t 2

= 1 2 × 1.6× 10 −19 ×3200 9.1× 10 −31 × ( 0.1 4× 10 7 ) 2 m

=01.76 mm

Hence, the correct answer is option: 'A'.

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