Question

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.

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)V_x will remain same

V_x=2*10⁶ m/s

Let us find V_y

V_y=U_y+at, U_y=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=(V_x²+V_y²)^(.5)

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