Q. An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity $10^6$ m/s. If the magnitude of the electric field between the plates is 9.1 V/cm, then the vertical component of velocity of electron is (mass of electron $=9.1\times10^{-31}$ kg and charge of electron $=1.6\times10^{-19}$ C)
A) 0
B) $1\times10^6$ m/s
C) $16\times10^6$ m/s
D) $16\times10^4$ m/s
Explanation
[Solution by Dr. Abubacker] $a=\frac{eE}{m}=\frac{1.6\times10^{-19}\times910}{9.1\times10^{-31}}=1.6\times10^{14}$ m/s². Time in the field $t=\frac{0.1}{10^6}=10^{-7}$ s, so $v_y=at=1.6\times10^7=16\times10^6$ m/s. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.