Question

An electron in a computer monitor enters midway between two parallel oppositely charged plates as shown below. The initial speed of the electron before it enters the plates horizontally is 6.15 x 10' m/s. The electron is deflected a vertical distance of 4.70 mm down by the time it exits the plates. What is the magnitude and direction of the electric field between the plates? 2. electron v 6.15 x 10 m/s 4.7 mm 10 cm

if anyone can explain this in algebra physics that'd be great. can't get the right answer. thanks

Answer 1

As it can be seen, an acceleration acts in a downward direction,

Horizontal Component of motion,

$V_{x}= \frac{x}{t}$

$6.15*10^{7}= \frac{0.1}{t}$

$t = 1.626*10^{-9} s$

Consider vertical component of motion,

Using Kinematics,

$s = ut +0.5 at^{2}$

$4.7*10^{-3} =0*t +0.5 *a*(1.626*10^{-9})^{2}$

$a = 3.555*10^{15}m/s^{2}$

The electrostatic force is responsible for this acceleration

$F_{e}= ma$

$q*E= ma$

$E= \frac{ma}{e}$

$E= \frac{(9.1*10^{-31})*(3.555*10^{15})}{1.6*10^{-19}}$

$E= 2.0219*10^{4}N/C$ (downwards)