An electron is projected with an initial velocity o = 3.4 x 107 m/s along the y-

Question

An electron is projected with an initial velocity o = 3.4 x 107 m/s along the y-axis, which is the center line between a pair of charged plates. The plates are 1.00 m long and are separated by 0.10 m. A uniform electric field E, in the positive x-direction, is presented between the plates. The magnitude of the acceleration of the electron is 8.9 x 1015 m/s2. The magnitude of the electric field is changed so that an electron, projected with the same initial velocity, barely clears the far edge of one of the plates. What is the new magnitude of the electric field?

Explanation / Answer

initial vertical velocity of the electron is v =3.4*10^7 m/s length of the each plate is y =1.00 m seperation between two plates is d =0.10 m acceleration of the electron when travel between plates is a =8.9*10^15 m/s^2 consider vertical motion y = vt+1/2 at^2   since there is no vertical acceleration therfore time t =y/v                       =1/3.4*10^7                       =0.2941*10^-7 s consider horizontal motion in horizontal case there is no initial velocity x =ut+1/2at^2    since u =0 =1/2 at^2 but seperation between two plates is d =0.1 m so the value for x is x =0.05 m because the y -axis is the central line of the two plates ___________________________________________________________________________ but electrical force F=e*E but we can write F=ma eE=ma a =eE/m but x =1/2at^2          =1/2(eE/m)t^2          =1/2 E (e/m)t^2..............(1) but the value of mass of the electron is m =9.1*10^-31 kg and charge of the electron is e =1.6*10^-19C from the equation (1) the required electric feild is E =2xm/et^2    =2(0.05)(9.1*10^-31) /1.6*10^-19 *(0.2941*10^-7 s)    =657 N/C                                      

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