An electron enters a parallel plate capacitor with horizontal speed u and is found to deflect by ang

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1. Electric Charges and Fields

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

A

$0.1$

B

$0.2$

C

$0.8$

D

$1.6$

(KVPY-2014)

Solution

(a)

Electron is subjected to electric force due to field of capacitor plates.

Time taken by $e^{-}$to cross region between plates $=t=\frac{x}{u}$

In this time acceleration of $e^{-}$in $y$ direction is $a_y=\frac{F}{m}=\frac{e E}{m}$

Deflection $y$ is then,

$y=u_y t+\frac{1}{2} a_y t^2$

$\Rightarrow \quad y=\frac{e E x^2}{2 m u^2}$

Now, $\tan \theta \propto y \propto \frac{1}{u^2}$

So, $\quad \frac{\tan \theta_2}{\tan \theta_1}=\frac{u_1^2}{u_2^2}$

Given that,

$\tan \theta_1=0.4, u_2=2 u_1$

$\therefore \quad \tan \theta_2=0.4\left(\frac{u_1^2}{\left(2 u_1\right)^2}\right)=\frac{0.4}{4}=0.1$

Standard 12

Physics

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