Electric Field Projectile Motion

AI Thread Summary
Electrons will exit the uniform electric field at an angle determined by the slope of their trajectory, which relates to their velocity at that point. The horizontal position is given by the equation x = V0*t, while the vertical position is described by y = -((e*E)/(2*m*V0^2))(x^2). The velocity parallel to the plates remains constant, while the velocity perpendicular to the plates can be derived from energy conservation principles. The angle of exit is therefore influenced by both the horizontal and vertical components of the velocity. Understanding these dynamics is crucial for predicting the behavior of electrons in an electric field.
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At what angle will the electrons leave the uniform electric field at the end of the parallel plates? Assume the plates are 5.4 cm long and E = 5.0 x 10^3 N/C. Ignore fringing of the field. (counterclockwise from the x-axis is positive)

I know the horizontal position is given by x=V0*t and
y= -((e*E)/(2*m*V0^2))(x^2) where e is charge of the electron and x is the horizontal position...
 

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physicsss said:
At what angle will the electrons leave
..relates to the slope of the tangent to their trajectory. [That was a hint.]
 
1. Direction in which it leaves is same as that of velocity at the point.

2. Velocity parallel to the plates will remain same.

3. Velocity perpendicular to the plates can be calculated using force of energy conservation.
 
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