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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 :
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If the initial horizontal speed is doubled,then tan will be :
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Verified Answer
The correct answer is:
$0.1$
$\mathrm{v}_{\mathrm{y}}=\mathrm{ay} \cdot \mathrm{t}$
$\mathrm{v}_{\mathrm{y}}=\frac{\mathrm{e} E}{\mathrm{~m}} \cdot \frac{\mathrm{l}}{\mathrm{u}}$
$\mathrm{v}_{\mathrm{x}}=\mathrm{u}$
$\tan \theta=\frac{\mathrm{e} \mathrm{El}}{\mathrm{mu}^{2}}$
$\frac{\tan \theta_{1}}{\tan \theta_{2}}=4$
$\tan \theta_{2}=\frac{\tan \theta_{1}}{4}=0.1$
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