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AN ELECTRON MOVING WITH A SPEED 5 X 10^(8) CM/SEC IS SHOT PARALLEL TO AN ELECTRIC FIELD OF STRENGTH 1 X 10^(3) N/C ARRANGED SO AS TO RETARD ITS MOTION. (A)HOW FAR WILL THE ELECTRON TRAVEL IN THE FIELD BEFORE COMING TO REST(MOMENTARILY)? (B)HOW MUCH TIME WILL ELAPSE? (C)IF THE ELECTRIC FIELD ENDS ABRUPTLY AFTER 0.8 CM, WHAT FRACTION OF ITS ENERGY WILL THE ELECTRON LOOSE IN TRAVERSING IT?

AN ELECTRON MOVING WITH A SPEED 
5 X 10^(8) CM/SEC IS SHOT PARALLEL TO AN ELECTRIC FIELD OF STRENGTH 1 X 10^(3) N/C ARRANGED SO AS TO RETARD ITS MOTION.
(A)HOW FAR WILL THE ELECTRON TRAVEL IN THE FIELD BEFORE COMING TO REST(MOMENTARILY)?
(B)HOW MUCH TIME WILL ELAPSE?
(C)IF THE ELECTRIC FIELD ENDS ABRUPTLY AFTER 0.8 CM, WHAT FRACTION OF ITS ENERGY WILL THE ELECTRON LOOSE IN TRAVERSING IT?

Grade:Select Grade

2 Answers

siddharth gupta
28 Points
6 years ago
Initial speed of electron=5 x 10cm/s=5 x 106 m/s 
electron is to be shot parallel to the electric field in the same direction as that of the field so that a retarding force acts on it in opposite direction and it is momentarily brought to rest.
retarding acceleration of the electron=(1.602 X 10-19 X 103)/9.1X10-31 kg=1.75 X 1014 m/s2
(A.)using third eqn of uniformly accelerated motion 
0=0.25 X 1014 - 3.5 X 1014s
s=25/350 m=1/14 m=7.1 cm.
(B.)using first eqn of motion
 0=u-at
u=at
t=(5X106)/(1.75 x 10 14)=2.85 X 10-8sec.
(C)ASSUME THE POTENTIAL AT INITIAL POSITION OF ELECTRON TO BE ZERO
POTENTIAL AT A DISTANCE OF 0.8 cm=-8V.
WORK DONE =ENERGY LOST=qV=8
8 X 1.60  X 10-19 J=12.8 x 10-19J
WITH REGARDS 
SIDDHARTH GUPTA
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Bharat Makkar
34 Points
6 years ago
thanks
 

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