- #1

Larissa McNeil

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## Homework Statement

An electron is projected with an initial speed v0 = 4.65×106 m/s into the uniform field between the parallel plates in (Figure 1). The direction of the field is vertically downward, and the field is zero except in the space between the two plates. The electron enters the field at a point midway between the plates. If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

There is a diagram showing the horizontal distance as 2.00cm and the vertical distance as 1.00cm but the electric is situated at half of the vertical distance so 0.5cm.

## Homework Equations

kinematics equation: y = vt + 0.5at^2

E=F/q=ma/q[/B]

## The Attempt at a Solution

I solved for the horizontal and vertical parts using kinematics.

For horizontal: 0.02=(4.65*10^6)t which equals 4*10^-9

For vertical: (2*0.005)/(4*10^-9)^2 which equals 6.25*10^14

Then I used E=F/q=ma/q:

(9.11*10^-31)(6.25*10^14)/(1.602*10^-19)=3.55*10^3

I've had friends look at my work and they agree with it, but mastering physics keep saying that my answer is wrong. It is due this Sunday and my professor does not like to be emailed. [/B]