Question

A uniform electric field $E$ is created between two parallel, 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 of the path of the electron as it comes out of the field.

• A. ${\tan }^{-1}\frac{ml}{E{v_0}^2}$
• B. ${\tan }^{-1}\frac{El}{em{v}_0}$
• C. ${\tan }^{-1}\frac{eEl}{m{v_0}^2}$
• D. ${\tan }^{-1}\frac{El}{m{v_0}^2}$

Answer 1

The correct option is C ${\tan }^{-1}\frac{eEl}{m{v_0}^2}$

Since, no field and hence no force act on the electron along X-direction.

Hence, $v_x$ remains constant.

Hence, time taken$=t=\frac{l}{v_x}=\frac{l}{v_o}$

And force on electron is $F=qE=eE$ along y-direction.

Acceleration of electron $=a=\frac{F}{m}=\frac{eE}{m}$ along Y-direction

Initial velocity-

$u_x=v_o$ and $u_y=0$

Final velocity while getting out -

$V-x=v_o$

$v_y=u_y+at=\frac{eEl}{m{V}_o}$

Now, angle is given by-

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

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

Answer-(C)