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Electrostatics- Firstname 13: A: An iron nucleus (Q- 4.16x101 c and M- 9.35x1026 kg)s initially moving horizontally with a speed V-2x10 m/s. Upon reaching point, it enters a region with a constant vertical electric field of 8000N/C. Which of the two paths does the charge follow? Explain Lastname 9 B: Determine the kinetic energy of the particle 3x10s later. Hint: the motion of this charge is qualitatively identical to projectile motion.

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Answer #1

13 (A) It follows path A. Because positive charge experiences force in the direction of electric field so it will gain velocity in vertical direction after entering in the electric field region.

(B)Vx will remain same

Vx=2*106 m/s

Let us find Vy

Vy=Uy+at, Uy=0, a=qE/m=((4.16*10-18)(8000)/(9.35*10-26))   ,t=(3*10-6)

Vy=1.07*106   m/s

speed=V=(Vx2+Vy2)(.5)

Kinetic energy=K.E.=(.5)mV2 =2.4 ×10−13     Joule

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