- #1

KillerZ

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## Homework Statement

An electron is launched at a 45 angle and a speed of 5.0*10^6 from the positive plate of the parallel-plate capacitor shown in the figure. The electron lands 4.0 cm away.

What is the electric field strength inside the capacitor?

What is the minimum spacing between the plates?

## Homework Equations

F = ma

F

_{on q}= qE

## The Attempt at a Solution

I am just wondering if this is correct?

V

_{o}= 5x10^6 m/s

V

_{oy}= (sin45)(5x10^6 m/s) = 3.5x10^6 m/s

t = [tex]\frac{(0.04m)(\sqrt{2})}{5x10^6 m/s}[/tex] = 11x10^-9 s

0 = V

_{oy}t + (1/2)at^2

a = [tex]\frac{-(3.5x10^6 m/s)(11x10^-9 s)}{(\frac{1}{2})(11x10^-9 s)^2}[/tex]

= -6.4x10^-14 m/s^2

F=ma

=(9.11x10^-31 kg)(-6.4x10^14 m/s^2) = -5.8x10^-16 N

F

_{on q}= qE

E = F/q = -5.8x10^-16 N / 1.60x10^-19 C

= -3.6x10^3 N/C

y

_{max}= V

_{0y}/(2a)

= 3.5x10^6 m/s / [(2)(-6.4x10^-14 m/s^2)]

= 2x10^-9 m