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tanya tandon Grade: 11
        A proton has an initial velocity of 3.82x10^7 m/s in the horizontal direction. It enters a uniform electric ?eld of 20000 N/C directed vertically. Ignoring gravitational e?ects, ?nd the time it takes the proton to travel 0.066 m horizontally. What is the vertical displacement of the pro- ton after the electric field acts on it for that time? What is the proton’s speed after being in the electric field for that time?
7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points
										

Dear tanya


since no force in the horizontal direction so horiizonta velocity will reamain constant Vx = 3.82x107 m/s


 


so time taken to travel 0.066 m horixontally is  t = .066/3.82x107 = 1.73  X 10-9 s


 


charge of the proton is q = 1.6 * 10-19C


mass  of the proton is m = 1.67 * 10-27 kg


given E =20000 N/C


 acceleration of proton in vertical direction  a=qE/m


                                                                               =1.916 * 1012  m/sec2


 


vertical distance moved by proton  h= 1/2  at2


                                                                   =1/2  * 1.916 * 1012 *(1.73  X 10-9)2


                                                                   =2.867 * 10-6 m


 vertical velocity of proton after time  t   Vy= at


                                                                            =1.916 * 1012 *(1.73  X 10-9)


                                                                             =3.315 * 103 m/sec


so resultanat velocity will be  = √Vx2 + Vy2


 


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Badiuddin


 

7 years ago
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