- #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
Fon q = qE
The Attempt at a Solution
I am just wondering if this is correct?
Vo = 5x10^6 m/s
Voy = (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 = Voyt + (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
Fon q = qE
E = F/q = -5.8x10^-16 N / 1.60x10^-19 C
= -3.6x10^3 N/C
ymax = V0y/(2a)
= 3.5x10^6 m/s / [(2)(-6.4x10^-14 m/s^2)]
= 2x10^-9 m