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?

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   V_y= 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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