# Homework Help: Electron entering capacitor

1. Jan 27, 2009

1. The problem statement, all variables and given/known data

The figure below shows an electron entering a parallel-plate capacitor with a speed of v=5.9e6 m/s. The electric field of the capacitor has deflected the electron downward by a distance of d=0.550cm at the point where the electron exits the capacitor.

Find the magnitude of the electric field in the capacitor.

Find the speed of the electron when it exits the capacitor.

2. Relevant equations

vf=vi+at
x=xi+vt
E=ma/q
t=sqrt(2h/a)

3. The attempt at a solution

x=xi+vt
0.00550m=0+5.9e6m/s+t
t=9.322e-10

t=sqrt(2h/a)
9.322e-10=sqrt((2 x 0.00550)/a)
a=5.17834e16

vf=vi+at
vf=5.9e6+(5.17834e16 x 9.322e-10)
vf=54172722.62

E=ma/q
E=(9.10938188 × 10-31 kg x 5.17834e16)/1.60217646 × 10-19 coulombs
E=294421.231

2. Jan 27, 2009

### Carid

This doesn't look right.
Minor error in second line of a plus sign where you should have a multiplication sign.
More serious error in that the longitudinal distance covered is what you should be speaking of here not the transverse distance due to the deflection.

Then when you've got the time for the electron to pass through the field you can figure that back in to work out the force causing the deflection.

3. Jan 27, 2009

### Astronuc

Staff Emeritus
Speed is simply the magnitude of velocity. Note that the electron enters the electric field with an initial velocity in the horizontal direction, transverse to the electric field. The electron is accelerated in the direction (vertical) of the electric field, not in the transverse direction.

Determine the transverse velocity component as the electron leaves the field, and find the speed based on the magnitudes of the two velocity components.