(6%) Problem 7: Consider the parallel-plate capacitor shown in the figure. The p

Question

(6%) Problem 7: Consider the parallel-plate capacitor shown in the figure. The plate separation is 5.9 mm and the the electric field inside is 15 N/C. An electron is positioned halfway between the plates and is given some initial velocity, v. Randomized Variables d=5.9mm E= 15N/C ©theexpertta.com 50% Part (a) What speed, in meters per second, must the electron have in order to make it to the negatively charged plate? Grade Summary 0% 100% sin) cotan) atan0 cosh) tnh cotanho tano acoso acotan sinh0 cosO Attempts remaining: 2 (1% per attempt) detailed view ODegrees O Radians Hint I give up Hints:--deduction per hant. Hints remaining:- Feedback:-deduction per feedback. 50% Part (b) If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. What w its speed be, in meters per second, when it reaches the positive plate in this case? All content © 2017 Expert TA, LLC

Explanation / Answer

work done by the potential difference to move the electron is W = q*dV = 1.6*10^-19*E*(d/2)= 1.6*10^-19*15*(5.9*10^-3)/2 = 7.08*10^-21 J

But according to work energy theorem

Work done on the electron = change in kinetic energy

q*dV = 0.5*m*v^2

7.08*10^-21 = 0.5*9.11*10^-31*v^2

v = 1.25*10^5 m/s

b) given that Vi = v/2 = (1.25*10^5)/2 = 6.25*10^4 m/sec

distance travelled by electron toward negative plate is d

v^2 - vi^2 = 2*a*s

accelaration is a = Fe/m = (q*E/m) = -(1.6*10^-19*15)/(9.11*10^-31) = -2.64*10^12 m/s^2

0 = (6.25*10^4)^2-(2.64*10^12*d)

d = 0.00147 m

to tal distance travelled is d' = 0.00147+(5.9*10^-3/2) = 0.00442 m

then

using v^2 - vi^2 = 2*a*S

v^2 - 0^2 = 2*2.64*10^12*0.00442

v = 1.53*10^5 m/sec

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