Question

A uniform electric field E = (8m/e) V/m is created between two parallel plates of length 1 m 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 2 m/s. The angle of the deviation (θ) of the path of the electron as it comes out of the field will be_______.

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

Answer 1

Answer: (B)

$tan^{-1}(2)$

Answer 2

Time taken by the electron to cross will be, $t=\frac{1}{v_x}=\frac{1}{2} s$

Acceleration of the electron along y axis

$a_g=\frac{F}{m}=\frac{Ee}{m}=\frac{8 m}{e}×\frac{e}{m}=8 m s^{-2}$

$v_y$ of the electron when it has crossed the capacitor will be,

$v_y=a_yt=8×\frac{1}{2}=4 m s^{-1 }$

Therefore,

$tanθ=\frac{v_y}{v_x}=\frac{4}{2}=2$

⇒ $θ=tan^{-1}2$