How long will it take to stop the electrons (d) 7.00 cm v N/C direction nad y ou

Question

How long will it take to stop the electrons (d) 7.00 cm v N/C direction nad y outward O points I previous Answers sercPg 15 Pro63 soln Each of the electrons in a particle beam has a kinetic energy of 1.00 x 10-17 (a) What is the magnitude of the uniform electric field (pointing in the direction of the electrons movement t will stop these electrons in a distance of 7.0 cm? NWC (b) How long will it take to stop the electrons? 4.18e-1 beginning and check each step carefully, ns Your response dimers significantly from the correct answer Rework your solution from the (c) After the electrons stop, the electric field will continue to act on them, causing the electrons to accelerate in a direction oppasne o the B at what rate? 1.57 14 V m/s My Previous Answers SercP9 15 P064 24 points Protons are projected with an initial speed vo 9,450 m/s into a region where a uniform electric field of magnitude E-300 Nyc is present (see figure below). The protons are a horizontal distance of 1.21 mm from the point where the protons are launched. Target

Explanation / Answer

F=ma

E=F/q

E = 892.9/q

Eq=ma

892.9/m=a

sqrt(2KE/m)=v

m = 9.1 × 10-31 kg

v = 4.688 x 106 m/s

d=(1/2)(a)(t2)+(vi)(t)

Where d is the distance the electrons will travel, a is the acceleration, t is the time travelled, and vi is the initial velocity of the electrons.

0=(1/2)(a)(t2)+(vi)(t)-d

0 = 0.5 * 9.8120 x 1032 *t2 + 4.688 x 106 t – 0.07

t = 1.195 x 10-17 s

t = 1.195 x 10-8 ns

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