An electron q=−1.6×10−19, me=9.1×10−31 kg is projected out along the + x axis with an initial speed of 2.0×106 m/s. It goes 20 cm and stops due to a uniform electric field in the region. Find the magnitude and direction of the field.
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a
50 N/C, +X direction
b
57 N/C, +X direction
c
60 N/C, -X direction
d
67 N/C, -X direction
answer is B.
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Detailed Solution
V2−u2=2as−2.0×1062=2a×20×10−2a=−1013 ms−2F=ma=9.1×10−31×−1013=9.1×10−18 NE=Fq=9.1×10−181.6×10−19=57 N/C approxdirection should be in +X direction