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## Homework Statement

An electron is launched at a α=38.5° angle and speed of 4.12×10^6 m/s from the positive plate of the parallel plate capacitor shown. If the electron lands d=3.03 cm away, what is the electric field strength inside the capacitor?

## Homework Equations

t = d/v

v0y = v*sin0

v0x = v*cos0

x(t) = v0t + 1/2at^2

## The Attempt at a Solution

So I first solved for v0y, since the electric field strength is perpendicular and only depends on vertical velocity.

v0y = 2.57 x 10^6

I then found time, after finding the hypotenuse of the distance

d = sqrt(0.0303^2 + 0.0303^2)

d = 0.04

t = d/v

t = 0.04 / 4.12 x 10^6

t = 1.04 x 10^-8

After this, I know I need to find acceleration using

x(t) = v0y*t + 1/2at^2

And this is where I have problems. I'm not exactly sure that the above procedure is correct, but assuming that it is, do I make the position zero? So is it:

v0y = 2.57 x 10^6

t = 1.04 x 10^-8

a = ?

x(t) = 0

I know that after this I just have to substitute into

Eq = ma

And solve for E, but I'm having trouble with solving for acceleration..please help!

Thank you very much!