- #1

KAT444

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## Homework Statement

An electron initially moves in a horizontal direction and has a kinetic energy of 2 x 10^3 electron volts when it is in the position shown above. It passes through a uniform electric field between two oppositely charged horizontal plates (region I) and a field free region (region II) before eventually striking a screen at a distance of .08 meters from the edge of the plates. The plates are .04 meters long and are separated from each other by a distance of .02 meters. The potential difference across the plates is 250 volts.

a. Calculate the initial speed of the electron as it enters region I.

b. Calculate the magnitude of the electric field E between the plates, and indicate its

direction

c. Calculate the magnitude of the electric force F acting on the electron while it is in

region I

## Homework Equations

KE=(1/2)mv^2

1 eV=1.6 x 10^-19 J

V=Ed

F=qE

mass of electron is 9.11 x 10^-31

charge of electron is 1.602 x 10^-19

## The Attempt at a Solution

I attempted all three parts of the solution, and am looking for clarification on my work.

a. First I changed 2 x 10^3 into 3.2 x 10^-16 J

KE=.5mv^2

3.2 x 10^-16 = .5(9.11 x 10^-31)(v^2)

**2.7 x 10^7 m/s=v**

b. V=Ed

250=E(.02)

**12500 V/m=E**

c. F=qE

F=(-1.602 x 10^-15)(12500)

**F=2 x 10^-15**

Any help anyone could give me would be great, especially if I did parts b and/or c wrong.