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Electrostatics question

2022 · 28 Jul · Shift 2 · Q49
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  5. /2022 · 28 Jul · Shift 2 · Q49

Electrostatics question

2022 · 28 Jul · Shift 2 · Q49

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A uniform electric field E=(8 m/e) V/m\mathrm{E}=(8 \mathrm{~m} / \mathrm{e}) \,\mathrm{V} / \mathrm{m}E=(8 m/e)V/m is created between two parallel plates of length 1 m1 \mathrm{~m}1 m as shown in figure, (where m=\mathrm{m}=m= mass of electron and e = charge of electron). An electron enters the field symmetrically between the plates with a speed of 2 m/s2 \mathrm{~m} / \mathrm{s}2 m/s. The angle of the deviation (θ)(\theta)(θ) of the path of the electron as it comes out of the field will be ‾\underline{\hspace{2cm}}​. JEE Main 2022 (Online) 28th July Evening Shift Physics - Electrostatics Question 109 English
  1. A
    tan⁡−1(4)\tan ^{-1}(4)tan−1(4)
  2. B
    tan⁡−1(2)\tan ^{-1}(2)tan−1(2)
  3. C
    tan⁡−1(13)\tan ^{-1}\left(\frac{1}{3}\right)tan−1(31​)
  4. D
    tan⁡−1(3)\tan ^{-1}(3)tan−1(3)
View written solutionFree

Correct answer: B

  1. Given data
  • Electric field: E=(8me) V/mE=\left(\frac{8m}{e}\right)\,\text{V/m}E=(e8m​)V/m
  • Length of plates: L=1 mL=1\,\text{m}L=1m
  • Initial speed of electron (horizontal): ux=2 m/su_x=2\,\text{m/s}ux​=2m/s

The electron enters symmetrically, so initial vertical velocity is zero.

  1. Force and acceleration on the electron

Magnitude of electric force on electron: F=eE=e(8me)=8mF=eE=e\left(\frac{8m}{e}\right)=8mF=eE=e(e8m​)=8m

Hence acceleration: ay=Fm=8mm=8 m/s2a_y=\frac{F}{m}=\frac{8m}{m}=8\,\text{m/s}^2ay​=mF​=m8m​=8m/s2

So the electron has vertical acceleration of magnitude ay=8 m/s2a_y=8\,\text{m/s}^2ay​=8m/s2

  1. Time spent inside the field

Horizontal motion is uniform, so t=Lux=12 st=\frac{L}{u_x}=\frac{1}{2}\,\text{s}t=ux​L​=21​s

  1. Velocity components at the exit

Horizontal velocity remains unchanged: vx=2 m/sv_x=2\,\text{m/s}vx​=2m/s

Vertical velocity gained: vy=ayt=8×12=4 m/sv_y=a_y t=8\times \frac{1}{2}=4\,\text{m/s}vy​=ay​t=8×21​=4m/s

  1. Angle of deviation

At the exit, the path makes angle θ\thetaθ with the original horizontal direction, so tan⁡θ=vyvx=42=2\tan\theta=\frac{v_y}{v_x}=\frac{4}{2}=2tanθ=vx​vy​​=24​=2

Therefore, θ=tan⁡−1(2)\theta=\tan^{-1}(2)θ=tan−1(2)

  1. Check with options
  • A: tan⁡−1(4)\tan^{-1}(4)tan−1(4)
  • B: tan⁡−1(2)\tan^{-1}(2)tan−1(2)
  • C: tan⁡−1(1/3)\tan^{-1}(1/3)tan−1(1/3)
  • D: tan⁡−1(3)\tan^{-1}(3)tan−1(3)

So the correct option is: tan⁡−1(2)\boxed{\tan^{-1}(2)}tan−1(2)​

Hence, Option B is correct.

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