The figure shows an electron passing between two charged metal plates that creat

Question

The figure shows an electron passing between two charged metal plates that create a 106 N/C vertical electric field perpendicular to the electron's original horizontal velocity. (These can be used to change the electron's direction, such as in an oscilloscope.) The initial speed of the electron is 3.1 times 10^6 m/s, and the horizontal distance it travels in the uniform field is 3.8 cm. What is its vertical deflection in m? What is the vertical component of its final velocity in m/s? At what angle does it exit in degrees Neglect any edge effects.

Explanation / Answer

step;1

Given that

electric field E=106 N/c

initial speed u=3.1*10^6 m/s

horizantal distance d=3.8 cm

step;3

now we find the vertical diffelection

the vertical deflection y=1.6*10^-19*106*(3.8*10^-2)^2/2*(3.1*10^6)^2*9.1*10^-31

=2449.0*10^-23/174.9*10^-19

=14*10^-4 m

now we find the finial velocity

finial velocity V=u+qEX/m*u

=3.1*10^6+1.6*10^-19*106*3.8*10^-2/9.1*10^-31*3.1*10^6

=3.1*10^6+644.5*10^-15/28.21*10^-25

=3.1*10^6+22846.5*10^6

=22849.6*10^6 m/s

now we find the angle

angle =tan^-1[22849.6/3.1]=89.9 degree

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