The second photo are the choices you can choose from (a) In the figure, a proton

Question

The second photo are the choices you can choose from (a) In the figure, a proton is fired with an initial speed (%) 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 ov 500 v around (use x-0 as the location of the negative plate). ii. What is the potential at that location? ii. What is the proton's speed when it reaches the 100V Vo m/s location? iv. If an electron is fired from the positive plate with an initial velocity that is 80 times Jaster than the initial velocity of the proton, how far from the OV plate will it stop? (b) In the figure to the right: i. What is the escape speed of the proton (middle charge)? (Assume the two outer charges are fixed in place) mm 5 mm ii. If the proton (middle charge) is launch with half of -2 nC the escape speed, how far away from its starting point will it stop and turn around? 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 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 moment of inertia of the rotating rod about its center and don't forget the rotational kinetic energy term (along with the ofer terms) when you set up your conservation of energy problem). (c)

Explanation / Answer

a)
i)
Initial velocity of the proton, u = 1.5 x 105 m/s
Electric force on the proton, F = qE
Where q is the charge of the proton and E is the electric field of the proton
E = V/d
= 500 / 0.1
= 5000 V/m
q = 1.6 x 10-19 C

Acceleration on proton = F/m, where m is the mass of the proton.
m = 1.673 x 10-27 kg
a = [(1.6 x 10-19) x 5000] / (1.673 x 10-27)
= 4.782 x 1011 m/s2.

The final velocity when the proton stops is, v = 0

Using the equation, v2 - u2 = 2as,
The distance, s = u2/2a
= [(1.5 x 105)2] / (2 x 4.782 x 1011)
= 0.0235 m
2.35 cm

Total distance from the left plate = 5 + 2.35 = 7.35 cm.

ii)
10 cm corresponds to 500 V
! cm corresponds to 50 V
7.35 cm corresponds to 7.35 x 50
= 367 V

iii)
100 V is at a distance of 2 cm from the left plate.
Proton started from 7.35 cm at rest (Initial velocity, u = 0)
Distance covered, d = 7.35 - 2 = 5.35 cm
Acceleration, a = 4.782 x 1011 m/s2.

Consider v be the final velocity.
Using the equation, v2 - u2 = 2ad,
v = SQRT[2ad]
= SQRT[2 x (4.782 x 1011) x 0.0535]
= 2.26 x 105 m/s

iv)
Initial velocity of the electron, u = 80 x 1.5 x 105 m/s
= 120 x 105 m/s
Acceleration for the electron, a = eE/me
= [(1.6 x 10-19) x 5000] / (9.1 x 10-31)
= 8.79 x 1014 m/s2.
Final velocity, v = 0

Using the equation v2 - u2 = 2as,
Distance traveled, s = u2/2a
[(120 x 105)2] / [2 x (8.79 x 1014)]
= 0.0819 m
= 8.19 cm
Distance from the left plate = 10 - 8.19
= 1.81 cm.

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