Question

An electron is shot horizontally (x direction) with velocity of 2.2x10^5 m/s into a region with a uniform electric field of 10000.0N/C in the positive Y direction. What is its position(x and y) in 1.25 seconds?

Answer 1

Given that

An electron is shot horizontally (x-direction) with velocityof (v_x) =2.2 x 10⁵ m/s

Here the electric field is along the positive y direction (E)=10000.0 N/C

Charge of the electron (e) =1.6*10^-19C

Mass of the electron (m) =9.1*10^-31kg

Time (t) = 1.25s

We know that the electric field is moving in the positive ydirection then the force will be along the negative ydirection

                    F_y = -e E

                     may= -eE

                       a_y = -eE/m

                            =- (1.6*10^-19C)(10000.0 N/C)/(9.1*10^-31kg)

                           = -1.75*10¹⁵m/s²

Then the motion along the vertical direction is

                 y =v_yt +(1/2)a_yt²                     =0 +(0.5)(-1.75*10¹⁵m/s²)(1.25)²

                    = -1.36*10¹⁵m

Then position of the electron in the x direction is

                x = v_xt

                   = (2.2 x 10⁵ m/s )(1.25s)

                   = 2.75*10⁵m