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JEE Main 2023 (25 Jan, Shift 1) · Motion of Charges in an Electric Field

Motion of Charges in an Electric Field question from JEE Main 2023 (25 Jan, Shift 1)

Q16JEE Main 2023 · 25 JAN · SHIFT 1Medium

Motion of Charges in an Electric Field

A uniform electric field of 10 N/C is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy 0.5 eV. The length of each plate is 10 cm. The angle (θ\theta) of deviation of the path of electron as it comes out of the field is ___________ (in degree).

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 45

  1. Entry speed: 12mv2=0.5 eV⇒v2=em\frac{1}{2}mv^2 = 0.5\text{ eV} \Rightarrow v^2 = \frac{e}{m} (in SI with ee in C).
  2. vy=at=eEm⋅Lvv_y = at = \frac{eE}{m}\cdot\frac{L}{v}, so tan⁡θ=vyv=eELmv2\tan\theta = \frac{v_y}{v} = \frac{eEL}{mv^2}.
  3. tan⁡θ=eELe=EL=10×0.1=1\tan\theta = \frac{eEL}{e} = EL = 10 \times 0.1 = 1
  4. So θ=45∘\theta = 45^\circ.
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