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/ LON-CAPA Mastery_2_TEX - O X + C A Secure | https://loncapa.science.uoit.ca/res/uoit/joe/Physics%2011/29%20-%20The%20Electric%20Field/Mastery_2_The_Electric_field.problemi L2 Question 2 ----------- V, 45o +++++++++++ In the figure above, an electron is released at an angle of 45 degrees from the parallel-plate capacitor's positive side. The distance between the plates is 4.45 cm and the electric field strength inside the capacitor is 3.65x104 N/C. If the electron avoids touching the negative plate, what is its maximum possible initial speed? click "Submit Answer" below to submit all 2 of your answers. Submit Answer Tries 0/10 A n. - B:- O Type here to search o e a ê A 9 P3 wa A k (* US 2018-01-30 1 A a E ENG &20 PMExplanation / Answer
Given theta = 45
maximum height h = 4.45 cm = 0.0445 m
electric field E = 3.65 *10^4 N/c
the electron experience a force of F = q *E
F = (-1.6 * 10^-19) * 3.65 * 10^4
F = - 5.84 * 10^-15 N (Downward direction)
acceleration 'a' is
a = F / m
a = - 5.84 * 10^-15 / (9.1* 10^-31)
a = -6.47 * 10^15 m/s^2
from projectile motion
maximum height h = (v0)^2 * sin^2(theta) / 2*a
0.0445 = ((v0)^2 * sin^2(45)) / (2 * 6.47 * 10^15)
v0 = 3.39 * 10^7 m/s
This is the maximum possible initial speed
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