Problem 17.38 A proton is traveling horizontally to the right at 4.20x 10^6 m/s.
ID: 1405208 • Letter: P
Question
Problem 17.38 A proton is traveling horizontally to the right at 4.20x 10^6 m/s. Part A Find (a)the magnitude and (b) direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 330 cm. Part B Counterclockwise from the left direction Part C How much time does it take the proton to stop after entering the field? Part D What minimum field ((a)magnitude and (b)direction would be needed to stop an electron under the conditions of part Part E counterclockwise from the left directionExplanation / Answer
a) Here Vi=4.20*10^6m/s , d= 0.033m , m = 1.6726*10^-27kg , q= 1.6022*10^-19C
We will use equation E= F/q ---------------(1)
where F is the force on proton , thus F = ma -------------(2)
Equating (1) and (2),
ma=qE E = ma/q -----------------(2)
First we find ‘a’ by using kinematic equation, Vf^2 + Vi^2 +2ad
0^2 = (4.20*10^6)^2 + 2*a*0.033
a= -2.67*10^14 m/s^2
From (2),
F= (1.6726*10^-27kg)*( -2.67*10^14m/s^2)= -4.5*10^-13N
from (1),
E= (-4.5*10^-13N)/( 1.6022*10^-19C ) = -2.8*10^6 N/C
Thus magnitude of E = 2.8*10^6 N/C
b) Direction of E must be opposite of the velocity i.e. horizontally left.
c) Use kinematic equation Vf=Vi+a*t
0m/s = 4.20*10^6m/s + (-2.67*10^14 m/s^2)* t t=> 1.57*10^-8 s
d) Here m= 9.1095*10^-31kg
Acceleration again will be same as part a)
hence
a= -2.67*10^14 m/s^2
From (2),
F= (9.1095*10^-31kg)*( -2.67*10^14m/s^2)= -2.43*10^-16 N
from (1),
E= (-2.43*10^-16 N)/( 1.6022*10^-19C ) = -1.52*10^-10 N/C
Thus magnitude of E = -1516.7 N/C
e) Direction of E must be in the direction of the velocity i.e. horizontally right