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 $10^6 \mathrm{~m} / \mathrm{s}$. If the magnitude of the electric field between the plates is $9.1 \mathrm{~V} / \mathrm{cm}$, then the vertical component of velocity of electron is (mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$ and charge of electron $=1.6 \times 10^{-19} \mathrm{C}$ )

• A

  0
• B

  $1 \times 10^6 \mathrm{~m} / \mathrm{s}$
• Correct answer
  C

  $16 \times 10^6 \mathrm{~m} / \mathrm{s}$
• D

  $16 \times 10^4 \mathrm{~m} / \mathrm{s}$

Solution

$\begin{aligned} & \Rightarrow \mathrm{t}=\frac{\ell}{\mathrm{V}_{\mathrm{x}}}=\frac{10 \times 10^{-2}}{10^6}=10^{-7} \\ & \Rightarrow \mathrm{~V}_{\mathrm{y}}=\mathrm{u}_{\mathrm{y}}+\mathrm{a}_{\mathrm{y}} \mathrm{t} \\ & \Rightarrow \mathrm{V}_{\mathrm{y}}=0+\frac{\mathrm{eE}}{\mathrm{m}} \times 10^{-7} \\ & \Rightarrow \mathrm{~V}_{\mathrm{y}}=\frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}} \times 9.1 \times 10^{-2} \times 10^{-7} \\ & \Rightarrow \mathrm{~V}_{\mathrm{y}}=16 \times 10^6\end{aligned}$

Hence, the correct answer is option: 'C'.