JEE Main 2022 · 28 Jul · Shift 2 · Physics Q10

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PhysicsSingle correct+4 / -1Question 10 of 90

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 ______.

Two horizontal plates 1 m long, the upper positive and the lower negative, field E downward; an electron enters midway at 2 m/s and leaves deflected upward at angle θ.

\tan^{-1}(4)

\tan^{-1}(2)

\tan^{-1}\left(\dfrac13\right)

\tan^{-1}(3)

Answer and worked solution

Official answer: B

Deflection. $a=\frac{eE}m=8$ m s $^{-2}$ for $t=\frac12$ s, so $v_y=4$ m/s against $v_x=2$ m/s.

$\theta=\tan^{-1}2$ . (B)

Topics:Motion in magnetic and electric fields Projectile motion

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JEE Main 2022 · 28 Jul · Shift 2, Physics question 10