Question

Electrostatics - -Firstname Lastname 13: A: An iron nucleus (Q-4.16x101 C and M- 9.35x1026 kgis initially moving horizontally with a speed Vo-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? Explai 9 B: Determine the kinetic energy of the particle 3x10°s later. Hint: the motion of this charge is qualitatively identical to projectile motion.

Answer 1

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According to question Q = 4.16 times 10^- 18 C, m = 9.35 times 10^- 26 kg since the charge will follow path A because direction of electric field in upwards since positive charge experience charges experience force in direction opposite to that of E According the question nucleus is positively charge so it will experience force in direction of E upward & follow path A Therefore F = 9 E according to question, t = 3 times 10^- 6s X component of velocity v_x = 2 times 10^- 6 m/s v_y = at = q, epsilon, m & t v_y = 1.07 times 10^6 m/s Therefore v = squareroot v_x^2 + v_y^2 = 2.27 times 10^6 m/s kinetic energy = 1/2 times 9.35 times 10^- 26 times (2.27 times 10^6)^2 = 2.41 times 10^- 13 J