- 56

- 0

**1. Homework Statement**

An electron is fired at 4.0x10

^{6}m/s horizontally between the parallel plates as shown, (see diagram) starting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 x10

^{2}N/C. The seperation of the plates is 2.0 cm

A) find the acceleration of the electron

b)find the horizontal distance traveled by the electron when it hits the plate

c) find the velocity of the electron as it strikes the plate

**2. Homework Equations**

a)ma=εq

b) Δd

_{y}=V1

_{y}ΔT + 1/2 aΔT

^{2}

ΔD

_{x}= V1

_{x}ΔT

c)V2

_{x}=√V1

_{x}+ 2aΔd

_{y}

**Really unsure about this, using x and y components in the same equation?**

q

_{electron}= -1.6x10

^{-19}

M

_{electron}= 9.11x10^ -31 kg

**3. The Attempt at a Solution**

A) ma=εq

a= (4.0x10^2)(1.6x10^-19)/(9.11x10^-31)

a= 7.03 x10^13 m/s^2 [down]

B)Δd

_{y}=V1

_{y}ΔT + 1/2 aΔT

^{2}

0.02=0ΔT +1/2 (7.03x10

^{13}ΔT

^{2}(the 0Δ T cancels?)

ΔT= √(0.02)(2)/(7.03x10

^{13}

ΔT= 2.3x10

^{-8}s

ΔD

_{x}= V1

_{x}ΔT

ΔD

_{x}= (4.0x10^6)*(2.3x10^-8)

=0.092 m

=9.2 cm

C) V2

_{x}=√V1

_{x}+ 2aΔd

_{y}

V2= √(4.06x10^6)+2(7.03x10^13)(0.02) <--(

**should I use horizontal or vertical distance here)**

V2= 4.34x10^6 m/s ( it gets faster which is a good sign lol)

My biggest confusion is when to use vertical distance 2 cm or horizontal distance 9.2 cm .

I took a lot of time to write this up so any help is appreciated immensely. I feel like I did everything right I just need re assurance. Im a perfectionist and I want to get full marks.

Sorry forgot to attach diagram, its crude but its exactly what is happening

Last edited: