An electron enters a parallel plate capacitor with horizontal speed $\mu$ and is found to deflect by angle $\Theta$ on leaving the capacitor as shown. It is found that $\tan \Theta=0.4$ and gravity is negligible
If the initial horizontal speed is doubled,then tan will be :

• Correct answer
  A

  $0.1$
• B

  $0.2$
• C

  $0.8$
• D

  $1.6$

Solution

$\mathrm{v}_{\mathrm{y}}=\mathrm{ay} \cdot \mathrm{t}$ $\mathrm{v}_{\mathrm{y}}=\frac{\mathrm{e} E}{\mathrm{~m}} \cdot \frac{\mathrm{l}}{\mathrm{u}}$ $\mathrm{v}_{\mathrm{x}}=\mathrm{u}$ $\tan \theta=\frac{\mathrm{e} \mathrm{El}}{\mathrm{mu}^{2}}$ $\frac{\tan \theta_{1}}{\tan \theta_{2}}=4$ $\tan \theta_{2}=\frac{\tan \theta_{1}}{4}=0.1$

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