Question

A uniform electric field $E$ is created between two parallel and oppositely charged plates as shown in figure. An electron enters the field symmetrically between the plates with a speed $v_0$. The length of each plate is $l$. Find the angle of deviation (which is given by $\theta$ in the figure) of the path of the electron as it comes out of the field interms of mass of electron $m$ and charge of the electron $e$. (Acceleration due to gravity can be neglected)

• A. $\theta ={\tan }^{-1}\left(\frac{2eEl}{m{v}_0^2}\right)$
• B. $\theta ={\tan }^{-1}\left(\frac{eEl}{m{v}_0^2}\right)$
• C. $\theta ={\cot }^{-1}\left(\frac{2eEl}{m{v}_0^2}\right)$
• D. $\theta ={\tan }^{-1}\left(\frac{eE}{m{v}_0^2}\right)$

Answer 1

The correct option is B $\theta ={\tan }^{-1}\left(\frac{eEl}{m{v}_0^2}\right)$

Here, the only force on the electron is due to the electric field. Hence, by using Newton's second law of motion.

$F=eE=ma$

$\Rightarrow a=\frac{eE}{m}$

So, the acceleration of the electron is $a=\frac{eE}{m}$ in the upward direction.

The horizontal velocity remains $v_0$ as there is no acceleration in this direction. Thus, the time taken to cross the plates is

$t=\frac{l}{v_0}.....\left(1\right)$

The upward component of the velocity of the electron as it emerges from the field region is

$v_y=u_y+at=0+\frac{eEl}{m{v}_0}$

The horizontal component of the velocity remains

$v_x=v_0$

The angle $\theta$ made by the resultant velocity with the original direction is given by

$\tan \theta =\frac{v_y}{v_x}=\frac{eEl}{m{v}_0^2}$

Thus, the electron deviates by an angle

$\theta ={\tan }^{-1}\frac{eEl}{m{v}_0^2}$

Hence, option (b) is correct answer.