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An electron moving with the speed 5 xx 1...

An electron moving with the speed `5 xx 10^6` m per second is shooted parallel to the electric field of intensity `1 xx 10^3` N/C. Field is responsible for the retardation of motion of electron. Now evaluate the distance travelled by the electron before coming to rest for an instant, (mass of `e = 9.1 xx 10^(-31)`kg)

A

0.7 cm

B

0.7 mm

C

7 m

D

7 cm

Text Solution

Verified by Experts

The correct Answer is:
D
Electric flux = NC force
`qE = mA`
`therefore a=(qE)/m =(1.6 xx 10^(-19) xx 10^(3))/(9.1 xx 10^(-31))`
`therefore a=0.1758 xx 10^(15) m//s^(2)`
Now equation of motion `v^(2) -v_(0)^(2) = 2ad`
`therefore d =(v^(2)-v_(0)^(2))/(2a)`, where `v=0, v_(0)=5 xx 10^(6) m//s`
`therefore =(25 xx 10^(12))/(3516 xx 10^(11)) = 0.00711 xx 10 = 0.0711` m
`=7.11 cm = 7 cm`
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