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

(C) $\tan ^{-1}\left(\frac{1}{3}\right)$

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Laws of Motion

A uniform electric field

E = (8 m / 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)✓ Correct
(C) $tan^{- 1} (\frac{1}{3})$
(D) $tan^{- 1} (3)$

Answer: (B)

Solution

The acceleration in the y-direction is

a_{y} = e E / m = e (8 m / e) / m = 8

m/s². Time to traverse the plates:

t = L / v_{x} = 1/2 = 0.5

s. Final y-velocity:

v_{y} = a_{y} t = 8 \times 0.5 = 4

m/s. The deflection angle:

tan θ = v_{y} / v_{x} = 4/2 = 2

, so

θ = tan^{- 1} (2)

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Single correct option

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