[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 FOR THE QUESTIONS ARE THE SECOND PICTURE, DON’T FORGET TO SHOW ALL WORK AND BE NEAT][:-):-)]

The Correct Answer Will be in the Multiple Choice options in the picture below to know if you did it right! It should help you a lot... :-)

500 V 4. 0 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. i. Find the location where the proton stops and turns (a) around (use x-0 as the location of the negative plate). i. What is the potential at that Vo m/s ii. What is the potential at that location? ii. What is the proton's speed when it reaches the 100V location? iv. If an electron is fired from the positive plate with an initial velocity that is 80 times faster than the initial velocity of the proton, how far from the 0V plate will it stop?

Explanation / Answer

Looks like the options provided are for plate with 2.5cm separation(mentioned in the image itself). I am solving the question per 10cm separation between plates since that is the seperation mentioned in the question :

The problem can be solved using energy conseravtion

mass of proton = 1.67*10-27kg

initial velocity of proton = 150000m/s

i) initial potential energy of proton = qV

Since proton is fired from midpoint, V =(500+0)/2 = 250V at midpoint

q = 1.6*10-19 C

therefore, total initial energy of proton = qV + (1/2)mv2

= 4*10-17 + 1.87875*10-17 = 5.87875*10-17 Joule

When the proton is stops, its velocity is zero

Let voltage at the point it stops be V

qV = 5.87875*10-17

V = 367.4Volts

we know that rate of increase of voltage with space = 500/10 = 50 V/cm

Therefore, the value of x coordinate at which voltage becomes 367.4V = 367.4/50 = 7.348cm

ii)As determined in part (i), the volatge at this point = 367.4 V

iii) Let the speed of proton at 100V location be v

5.87875*10-17 = (1.6*10-19)*(100) + (1/2)*(1.67*10-27)v2

v =2.26*105 m/s

iv)intiial velocity of electron = 80*150000 = 12000000m/s

mass of electron = 9.11*10-31 kg

initial potential = 500V

total initial energy of electron = (-1.6*10-19*500)+(0.5*9.11*10-31*120000002)

= -8*10-17+6.559*10-17 = -1.44*10-17 Joule

when the elctron stops,cthe entire energy is converted into potential energy

Let the voltage at that point be V

qV = -1.44*10-17

V = 90V

Therefore, distance from 0 V plate at which the electron stops = 90/50 = 1.8 cm

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