o V 500 v In the figure, a proton is fired with an initial speed (Vo) of 150,000
ID: 1731141 • Letter: O
Question
o V 500 v 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. (a) Find the location where the proton stops and turns around (use x-0 as the location of the negative plate). i. Vo m/s ii. What is the potential at that location? ii. What is the proton's speed when it reaches the 100V iv. If an electron is fired from the positive plate with an initial velocity that is 80 times location? faster than the initial velocity of the proto n, how far from the 0V plate will it stop? (b) In the figure to the right: -2 nC i. What is the escape speed of the proton (middle charge)? (Assume the two outer charges are fixed in place) 5mm1 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? (c) An electric dipole consists of 1.0 g sphess charged to +3.0nC at the ends of a 15 cm long non-conducting rod of mass 4g (don' 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 Ijust 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 other terms) when you set up your conservation of energy problem)Explanation / Answer
answering all parts of (a)
We will use conservation of energy:
i) Let the poiint be "x" m from the negative plate,
initial eneergy of proton = KE + PE = 1/2 m vi^2 + qV/2
Electric filed = V/d = 500 / (0.1) = 5000 N/C
plugging the values to find the initial KE =0.5 ( 1.67x 10^-27) ( 1.5 x 10^5) ^2 + 1.6 x 10^-19 ( 250) = 5.878 x 10^-17 J
The point at which proton rebounds will have only PE and no KE= q ( V/x) = q ( Ex)
5.878 x 10^-17 = 1.6 x 10^-19 (5000 x)
5.878 = 1.6X 50(x)
x = 7.347 cm from the positive plate
ii) V = Ed= 5000 ( 7.347 / 100) =367.375 V
iii) Again we will use conservation of energy,
Initial Eneergy at 250 V = Final energy at 100 V
5.878 x 10^-17 J= 1/2 ( 1.67x 10^-27) vf^2 + 1.67 x 10^-19 ( 100)
5.878 = 0.835 x 10^-10 vf^2 + 1.67
vf^2= 5.039 x 10^ 10
vf= 2.245 x 10^5 m/s apprx
iv)
F on electron = Eq
retardation = Eq/ m = 5000 ( 1.6 x 10^-19 ) / ( 9.109 x 10^-31) = 878.252 x 10^ 12 m/s^2
vf^2= vi^2 - 2as
0 = (1.5 x 80 x 10^ 5)^2 - 2 ( 878.252 x 10^ 12 ) (x)
( 878.252 ) (x) = 72
x= 1.8 cm apprx from the positive plate
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