An electron is moving east in a uniform electric field of 1.49 N/C directed to t
ID: 1540010 • Letter: A
Question
An electron is moving east in a uniform electric field of 1.49 N/C directed to the west At point A, the velocity of the electron is 4.45 times 10^5 m/s pointed toward the east. What is the speed of the electron when it reaches point B, which is a distance of 0.400 m east of point A? upsilon_e = 6.38 times 10^5 m/s A proton is moving in the uniform electric field of part A. At point A. the velocity of the proton is 1.93 times 10^4 m/s, again pointed towards the east What is the speed of the proton at point B? upsilon_p =Explanation / Answer
An electron is moving east in a uniform electric field of magnitude 1.54 N/C directed to the west.
Force on electron will be directed to east and acceleration of electron will be positive
Initial speed = u = 4.45×10^5 m/s
Acceleration = a =F/m=qE/m
Acceleration = a =1.6*10^-19*1.54 /9.11*10^-31
Acceleration = a = 2.7047*10^11 m/s^2
distance = s = 0.400 m
Using v^2 - u^2 = 2as
v = sq rt [ u^2 +2as]
the speed of the electron when it reaches point B= v = sq rt [u^2 +2as]
v = sq rt [(4.45*10^5)^2 + 2* 2.7047*10^11* 0.400 ] =1.98025+2.16376
v = 3.57097 *10^5 m/s
The speed of the electron when it reaches point B is 3.57097 *10^5 m/s
B) A proton is moving in the uniform electric field .At point A, the velocity of the proton is 1.88×10^(4) m/s, again pointed towards the east.
Initial velocity of proton = u =1.88×10^4 m/s
The proton is retarted because force of field is west and initial velocity is east.
the speed of the proton at point B = v = sq rt [ u^2 +2as]
acceleration of proton = a = -F/m=-qE/m=-1.6*10^-19*1.54/1.67*10^-27
acceleration of proton = a = -1.4754*10^8 m/s^2
the speed of the proton at point B = v = sq rt [ (1.88×10^4 )^2 -2*1.4754*10^8*0.400]
the speed of the proton at point B = 1.5343*10^4 m/s
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