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Home/ Questions/Q 7173
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Ashok0210
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Ashok0210
Asked: February 13, 20212021-02-13T07:35:19+00:00 2021-02-13T07:35:19+00:00In: Class 12 Physics

Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vₓ = 2.0 × 10⁶ m s⁻¹. If E between the plates separated by 0.5 cm is 9.1 × 10² N/C, where will the electron strike the upper plate? (|e|=1.6 × 10⁻¹⁹ C, mₑ = 9.1 × 10 ⁻³¹kg.)

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Class-12th Physics, CBSE and UP Board
Electric Charges and Fields,
Chapter-1, Exercise -1, Additional Exercise, Q-1.34
NCERT Solutions for Class 12th Physics

2020-2021cbsechapter 1class 12 physicsclass 12thclass 12th scienceelectric charges and fieldsncertncert solutionsphysics
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    1. Pawan1308
      2021-02-18T15:28:08+00:00Added an answer on February 18, 2021 at 3:28 pm

      Velocity of the particle, Vx = 2.0 x 106 m/s

      Separation of the two plates, d = 0.5 cm = 0.005 m

      Electric field between the two plates, E = 9.1 x 102 N/C

      Charge on an electron, q = 1.6 x 10–19 C

      Mass of an electron, me= 9.1 x 10-31 kg

      Let the electron strike the upper plate at the end of plate L, when deflection is s. Therefore,

      s= (qEL2)/(2mV2x)

      =>L = √ (2smV2x)/qE

      =  √ 2 x 0.005 x 9.1 x 10-31)/(1.6 x 10–19 x 9.1 x 102

      = √ 0.00025  = 0.016 m=1.6 cm

      Therefore ,the electron will strike the upper plate after travelling 1.6cm.

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