An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of $10\ \text{cm}$ length. The electron emerges out of the electric field region with a horizontal component of velocity $10^6\ \text{m/s}$. If the magnitude of the electric field between the plates is $9.1\ \text{V/cm}$, then the vertical component of velocity of electron is (mass of electron $= 9.1\times10^{-31}\ \text{kg}$ and charge of electron $= 1.6\times10^{-19}\ \text{C}$)
Answer: (C) $16\times10^{6}\ \text{m/s}$
The horizontal velocity is unchanged, so the time spent between the plates is
$$t = \frac{L}{v_x} = \frac{0.10}{10^6} = 10^{-7}\ \text{s}$$
The field is $E = 9.1\ \text{V/cm} = 910\ \text{V/m}$, so the vertical acceleration is
$$a = \frac{eE}{m} = \frac{1.6\times10^{-19}\times 910}{9.1\times10^{-31}} = 1.6\times10^{14}\ \text{m/s}^2$$
Vertical velocity on leaving the plates:
$$v_y = at = 1.6\times10^{14}\times10^{-7} = 1.6\times10^{7} = 16\times10^{6}\ \text{m/s}$$
Solution by Sreeraj P, M.Sc Physics