An electron Is projected with an initial speed v_0 = 1.25 times 10 degree m/s in

Question

An electron Is projected with an initial speed v_0 = 1.25 times 10 degree m/s into the uniform field between the parallel plates in the following figure. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. (Assume d_1 = 2.75 cm and d_2 = 5.50 cm.) (a) If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field. N/C (b) Suppose that in the figure the electron is replaced by a proton with the same initial speed vq. Would the proton hit one of the plates? Yes No If the proton would not hit one of the plates, what would be the magnitude and direction of its vertical displacement as it exits the region between the plates? (Enter 0 m and 'no direction' if the proton hits one of the plates.)

Explanation / Answer

Apply kinematic equation

s= vot + 1/2 at^2

a=0 becuase horizontal accleratrion is zero

s= vo t

t= s/ vo = 0.055 m/1.25 * 10^6 m/s = 4.4 * 10^-8 s

The equaiton for vertical displacment of electron is

y = voy t + 1/2 ay t^2

= 0 + 1/2 ay t^2

ay = 2y/ t^2

= 2( 2.75 * 10^-2 m/2)/(4.4 * 10^-8 s)^2

=0.142 * 1014 m/s^2

F = Eq

ma = Eq

E = ma/ q

=9.11 * 10^-31 ( 0.14 * 10^14)/1.6 * 10^-19

=80.87 N/C

(b)

a= Eq/m

= 80.87 N/C ( 1.6 * 10^-19)/1.673 * 10^-27

=77.34 * 10^8 m/s^2

y = 1/2 at^2

= 1/2 ( 77.34 * 10^8 m/s^2) ( 4.4 * 10^-8 s)^2

=7.48 * 10^-6 m

direction

theta = tan^-1 ( 7.48 * 10^-6 /0.055) = 7.79 * 10^-3 below horizontal

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.