Question

A uniform electric field $E=(8m/e)V/m$ is created between two parallel plates of length $1m$ as shown in figure, (where $m=$ mass of electron and $e=$ charge of electron). An electron enters the field symmetrically between the plates with a speed of $2m/s$. The angle of the deviation $(\theta )$ of the path of the electron as it comes out of the field will be

• A. ${\tan }^{-1}$ (4)
• B. ${\tan }^{-1}$ (2)
• C. ${\tan }^{-1}\left(\frac{1}{3}\right)$
• D. ${\tan }^{-1}$ (3)

Answer 1

Answer: (B)

$a_y=\frac{F_y}{m}=\frac{e(E)}{m}=\frac{e\left(\frac{8m}{e}\right)}{m}=8m/s^2$
$s_x=u_{x}t$
$1=2\times t$
$t=\frac{1}{2}\sec$
$v_y=u_y+a_{y}t$
$v_y=0+8\times \frac{1}{2}$
$v_y=4m/s$
$\tan \theta =\frac{v_y}{v_x}=\frac{4}{2}=2$
$\Rightarrow \theta ={\tan }^{-1}(2)$