Answer to Question #27125 in Electricity and Magnetism for Sloan Jenkins

Question

A proton with an initial horizontal velocity of 2.3X10^6 m/s enters a region of uniform electric field of strength 1.1X10^5 N/C between two oppositely charged parallel plates (top plate=negative, bottom plate=positive). The plates are 1.3 cm apart and 6.7 cm long. Calculate the distance below the top plate the particle must be shot so as to just miss hitting the right edge of the top plate. Ignore gravitational forces. m(sub)p=1.67X10^-27

Answer: 0.0045 m

Answer 1

First we will findacceleration of the proton between those plates. Force acting onproton can be found as
F = e*E= 1.6*10^(-19) * 1.1*10^5 N/C = 1.76*10^(-14) N
The acceleration thenis
a = F/m_p= 1.76*10^(-14) N/1.67*10^(-27) = 1.05*10^(13) m/s^2
Now we can write thelaw of motion for proton in Y axis (perpendicular to plates)
s=at^2/2
where sis distance to the top plate at which the particle must be shot so as to just miss hitting the right edge of the top plate. To find s, we need t,time of motion between the plates, which can be found from law of motion in X axis
l=vt
where l is length ofplates and v is horizontal velocity. From this
t= l/v =6.7*10^(-2)/(2.3*10^6) = 2.9*10^(-8) s
Now we find s
s=1.05*10^(13)*( 2.9*10^(-8))^2/2= 4.41*10^(-3) m = 0.441 cm