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Question

An electron moving with a speed of 5×106 ms1 is shot parallel to an electric field of strength 1000 Vm1, arranged so as to retard its motion. The electron travels a distance s in the field before coming momentarily to rest in time t. It is observed that the electric field ends abruptly after 0.85 cm and due to this the electron loses a percentage fraction f of its initial energy, then

A
s = 0.14 m
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B
s = 0.01 m
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C
t=0.09μs
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D
f=11%
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Solution

The correct option is D f=11%
Retardation a=forcemass=eEm
a=1.6×1019×10009×1031=1.78×1014 ms2
Since, v2v20=2as
02(5×106)2=2(1.78×1014)s
s=0.07m
Now, v=v0at
0=5×106(1.78×1014)t
t=2.8×108s=0.03 μs
Loss of energy = work done W = Fd = (eE)d
Fraction f=eEd12mv2=2eEdmv2
% Fraction=2×1.616×1000×0.8×1029×1031×(5×106)2=11%

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