(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant equations

F = ma

F_{on q}= qE

3. 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

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# Homework Help: An electron moving across capacitor problem

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