[PLEASE SHOW ALL WORK SO I CAN UNDERSTAND/LEARN][THERE ARE MULTIPLE CHOICE ANSWE

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[PLEASE SHOW ALL WORK SO I CAN UNDERSTAND/LEARN][THERE ARE MULTIPLE CHOICE ANSWERS FOR THE QUESTIONS SO YOU KNOW IF YOU ARE ON THE RIGHT TRACK. THEREFORE, IF YOU CALCULATE ONE OF THE NUMBERS THAT YOU SEE IN THE M/C ANSWERS THAN YOU ARE CORRECT SO IT WILL HELP GUIDE YOU][THE MULTPLE CHOICE ANSWERS ARE THE SECOND PICTURE, DON’T FORGET TO SHOW ALL WORK AND BE NEAT][:-):-)]

500 V 4. 0 V (a) In the figure, a proton is fired with an initial speed (vo) of 150,000 m/s from the midpoint of the capacitor toward the positive plate as shown. The plates are 10 cm. apart. i. Find the location where the proton stops and turns around (use x-0 as the location of the negative plate). What is the potential at that location? What is the proton's speed when it reaches the 100V ii. vo m/s iti. location? If an electron is fired from the positive plate with an initial velocity that is 80 times iv. faster than the initial velocity of the proton, how far from the 0V plate will it stop? (b) In the figure to the right: What is the escape speed of the proton (middle charge)? (Assume the two outer charges are fixed in place) i. 5 mm 5 mm ii. If the proton (middle charge) is launch with half of the escape speed, how far away from its starting point will it stop and turn around? -2 nC An electric dipole consists of 1.0 g spheres charged to ± 3.0 nC at the ends of a 15 cm long non-conducting rod of mass 4g (don't try to look up "electric dipole", it won't help (c) you. For the sake of this problem, it's just a fancy word for what I just described). The dipole rotates on a frictionless pivot at its center. The dipole is held perpendicular to a uniform electric field with a field strength 1000 V/m, then released. What is the dipole's angular velocity at the instant it is aligned with the electric field? (HINT: Look up the is the dipoles moment of inertia of the rotating rod about its center and don't ent of inertia of the rotating rod about its center and don't forget the rotational kinetic energy term (along with the other terms) when you set up your conservation of energy problem).

Explanation / Answer

Let x be the distance from where protons turns,

Vx = 500 x/ (0.025) = 5000 x

From conservation of energy

1/2 m v^2 = q V

1/2 x 1.67 x 10^-27 x (1.5 x 10^5)^2 = 1.6 x 10^-19 x 5000 x

x = 1.88 x 10^-17 / 1.6 x 10^-19 x 500 = 0.023 m

Hence, x = 0.023 m = 2.3 cm

ii)Vx = 5000 x 0.023 = 115 Volts

Hence, Vx = 115 Volts

iii)from conservation of energy

1/2 m v^2 = 1/2 m vf^2 - qV

vf = sqrt (v^2 + 2 qV/m)

vf = sqrt [(1.5 x 10^5)^2 + 2 x 1.6 x 10^-19 x 100/(1.67 x 10^-27)]

Vf = 2.04 x 10^5 m/s

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