(adsbygoogle = window.adsbygoogle || []).push({}); 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

Tried this problem several times but am not getting the right answer. I have 10 tries, but the computer keeps saying my answer is wrong. Please help.

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# Homework Help: Electron entering capacitor

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