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Q.

An electron moving with the speed 5×106 per sec is shooted parallel to the electric field of intensity 1×103N​/​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=9×10−31Kg. charge =1.6×10−19C)

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a

7 m

b

0.7 mm

c

7 cm

d

0.7 cm

answer is C.

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Detailed Solution

Electric force qE=ma⇒a=QEm∴a=1.6×10−19×1×1039×10−31=1.69×1015u=5×106 and v=0∴ From v2=u2−2as⇒s=u22a∴ Distance s=5×1062×92×1.6×1015=7cm.(approx)
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