Math, asked by aarushiaru2728, 1 month ago

AIRT EL
Example 1.8 An electron falls through a distance of 1.5 cm in a
uniform electric field of magnitude 2.0 x 104N C- [Fig. 1.13(a)]. The
direction of the field is reversed keeping its magnitude unchanged
and a proton falls through the same distance [Fig. 1.13(b)]. Compute.
the time of fall in each case. Contrast the situation with that of 'free
fall under gravity
te​

Answers

Answered by abhijitdey16122004
2

Answer:

ITS EASY

Step-by-step explanation:

S=1.5cm=1.5×10  

−2

m

             u=0m/s

m  

e

​  

 : mass of electron =9.10×10  

−31

Kg

m  

p

​  

 : mass of proton =1.672×10  

−27

Kg

from coloumb's force,  

F

=q  

E

 

and given  

E

=2×10  

4

N/c

force proton  

F

 

(+)

​  

=1.6×10  

−19

×2×10  

4

=3.2×10  

−19

N

force on electron,  

F

 

(−)

​  

=−1.6×10  

−19

×2×10  

4

=−3.2×10  

−15

N

From Newton's Second law,    

F

=m  

a

 

∴  for proton, a  

(+)

​  

=  

m  

p

​  

 

F

 

(+)

​  

 

​  

=1.913×10  

12

m/s  

2

 

for electron, a  

(−)

​  

=  

m  

e

​  

 

F

 

(−)

​  

 

​  

=−0.35×10  

16

m/s  

2

 

from equ of motion, S=ut+  

2

1

​  

at  

2

⟶(1)

ATQ, The direction of the field was reversed, hence we will ignore the sign convention for acceleration.

∴  Substituting values in equ (1) for proton,

1.5×10  

−2

=  

2

1

​  

×1.913×10  

12

×t  

2

 

1.568×10  

−14

=t  

2

 

⇒t=1.252×10  

−7

seconds

Substituting values in equ(1) for electron,

1.5×10  

−2

=  

2

1

​  

×3.5×10  

15

×t  

2

 

0.857×10  

−17

=t  

2

 

⇒t  

0

​  

=2.927×10  

−19

seconds

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