Trajectory of a Charge in an Electric Field

[TheLegace]

Homework Statement

An charge with mass m and charge q is emitted from the origin, (x=0,y=0). A large, flat screen is located at x=L. There is a target on the screen at y position yh, where yh>0. In this problem, you will examine two different ways that the charge might hit the target. Ignore gravity in this problem.

Assume that the charge is emitted with velocity v0 in the positive x direction. Between the origin and the screen, the charge travels through a constant electric field pointing in the positive y direction. What should the magnitude E of the electric field be if the charge is to hit the target on the screen?

Homework Equations

The equations and questions is from this thread: https://www.physicsforums.com/showthread.php?t=59814

I just don't have much of an idea of where the y_h equation is coming from. Is from the forces or kinematics. I figured this equation is coming from original y = 1/2at^2, so I am assuming it is coming from a Newtons Law equation. F=ma=Fq. If not could someone please help, as soon I get to the y_h formula I think I can figure out the rest.

The Attempt at a Solution

x(t)=v0*t
y(t)=(1/2)*(q*E/m)*t^2
tfinal= L/v0

Thank You.

 

[LowlyPion]

I just don't have much of an idea of where the y_h equation is coming from. Is from the forces or kinematics. I figured this equation is coming from original y = 1/2at^2, so I am assuming it is coming from a Newtons Law equation. F=ma=Fq. If not could someone please help, as soon I get to the y_h formula I think I can figure out the rest.

The Attempt at a Solution

x(t)=v0*t
y(t)=(1/2)*(q*E/m)*t^2
tfinal= L/v0

Thank You.

You basically have it.

a comes from the q*E/m and you are using the acceleration, time distance relationship.

Plugging in L/V_o for t gives you the elevation y=h on the screen.